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The reaction of C6H5CH3 with O2 produces C6H5COOH and H2O.? The reaction of C6H5CH3 with O2 produces C6H5COOH and H2O. How many grams of the sample (C6H5CH3) are neede to produce 1.21 grams C6H5COOH? If 1.00 grams sample is burned with an excess air, how many moles of C6H5COOH are formed? If the reaction produced 62.40% more amount of product (C6H5COOH), how many grams of C6H5CH3 had reacted? How many grams O2 were consumed in the reaction?

George Carty replied: "Balanced chemical equation: 2C6H5CH3 + 3O2 --> 2C6H5COOH + 2H2O Toluene has a molecular weight of 92, while benzoic acid is 122, so 0.912 grams of toluene is needed. One gram of toluene makes 1.326 grams (0.01086 moles) of benzoic acid. if 62.40% more product then the same increase in reactant (1.624 grams toluene). This requires 0.847 grams of oxygen."

nevermoo replied: "Mr of toluene (TOL)= 92 Mr of benzoic acid (BA) = 122 No. of moles of BA = 1.21/122 = 0.009918 Mole ratio: 1 tol : 1 BA No. of moles of tol = 0.009918 (a) Mass of tol = 0.912g No. of moles of tol = 0.01087 Mole ratio: 1 tol : 1 BA (b) Mass of BA formed = 1.33g Mass of BA = 1.33g X 162.40% = 2.16g Mole ratio: 1 tol: 1 BA : 3/2 O2 No. of moles of BA = 2.16/122 = 0.0177 No. of moles of O2 = 0.0177 X 1.5 = 0.02655 No. of moles of tol = 0.0177 (c) Mass of toluene = 0.0177 X 92 = 1.63g Mass of O2 consumed = 0.02655 X 32 = 0.85g Usually the synthesis of benzoic acid from toluene requires the use of hot acidified potassium permanganate under reflux."

Sodium benzoate, C6H5COONa, is the salt of a weak avid, benzoic acid, C6H5COOH. A 0.10 molar solution of? sodium benzoate has a pH of 8.60 at room temperature. A. Calculate the [OH-] in the sodium benzoate solution described above. B. Calculate the value for the equilibrium constant for the reaction: C6H5COO- + H2O <--> C6H5COOH + OH- C. Calculate the value of Ka, the avid dissociation constant for benzoic acid. I'm not really sure what to do.. So, for A, is the equation x^(2) over 0.10 - x? For B, I think the equation is: [C6H5COOH] [OH] over [C6H5COO-] But I don't know how to solve for its value. Yeah, I don't know what I'm doing, could you help me? oops, I mean.. can*

vv replied: "The [OH]- by default would be calculated from your given pH by 14.0-8.60 = pOH = 5.40 5.40 = log 1 - log [OH]^- : log [OH]^- = 5.4 + 0 [OH]- = 4 x 10^-6 Your B is incomplete, but with all the conc etc your right, the Equil Constant would be the products of your Equil over the reactants (left side of equation); to evaluate; you have written the classical backword reaction.More appropriately, C6H5CO2H ------> C6H5CO2- + H+and in that case, mathematically that would be Keq = (x)^2/1-x but ionization so low that 1-x really = x. C.. The calculation would be: pH = pKa + log salt/acid (concentrations) you have pH + the conc. This gives you pK; and pK = log 1/K"

Help: Sodium benzoate, C6H5COONa, is the salt of a weak avid, benzoic acid, C6H5COOH. A 0.10 molar solution of sodium benzoate has a pH of 8.60 at room temperature. A. Calculate the [OH-] in the sodium benzoate solution described above. B. Calculate the value for the equilibrium constant for the reaction: C6H5COO- + H2O <--> C6H5COOH + OH- C. Calculate the value of Ka, the avid dissociation constant for benzoic acid. I'm not really sure what to do.. So, for A, is the equation x^(2) over 0.10 - x? For B, I think the equation is: [C6H5COOH] [OH] over [C6H5COO-] But I don't know how to solve for its value. Yeah, I don't know what I'm doing, can you help me? I asked this question earlier, but I didn't get any answers.

Dr.A replied: "C6H5COO- + H2O <----> C6H5COOH + OH- pOH = 14 - 8.60 =5.40 [OH-] = 10^-5.40 = 3.98 x 10^-6 M K = ( 3.98 x 10^-6)^2 / 0.10 =1.58 x 10^-10"

Which of the following reactions would yield benzaldehyde? C6H5CH(OCH3)2 --- (H3O+) ---> C6H5COOH --- (1. LiAlH4, ether; 2. H2O) ---> I know the answer is the first one, but I don't understand why. Could someone please explain both reactions to me? Thanks very much!

e^x replied: "I'm not sure exactly how the first one reaches benzaldehyde but I can tell you that the 2nd one does not work because LiAlH4 is a very strong reducing agent. LiAlH4 will reduce the carboxylic acid down to the carbonyl then down to the alcohol."

How do you find what the density of molecule is? I need it for Benzoic acid (C6H5COOH), Acetanilde (C6H5HNCOCH3), NaCl, HCl, NaOH, and H2O. I'm not sure how to find it, so if you could explain how to get the answer, so if you could explain it instead of just giving the answer it would be great. Thanks

Top Gun replied: "Look it up on Wikipedia"

A.V.R. replied: "The concept of density is not applicable to single molecules. If it is bulk material that is being discussed the answer is the trivial one of dividing the mass by the volume of the material. Depending on the state of the material (solid, liquid, or gas) the method will vary."

Show how acids ionize in an aqueous solution? Show how each of the following acids ionize in aqueous solution. 1. HI 2. HF 3. C6H5COOH I know that the beginning of each equation should look like this: 1. HI (aq) + H20 (l) -> ? 2. HF (aq) + H2O (l) -> ? 3. C6H5COOH (aq) + H2O (l) -> ? can anybody offer some insight??? Im not sure where to go from here...

Dr.A replied: "HI (aq) + H2O (l) ---> H3O+ (aq) + I- (aq) HF (aq) + H2O (l) ----> H3O+ (aq) + F- (aq) C6H5COOH (aq) + H2O (l) ----> C6H5COO- (aq) + H3O+ (aq)"

bmanhat replied: "By definition acids are H+ donors so in water they will release an H+ and form the conjugate base and the acid h3O"

percent ionized!? Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is a preservative found in foods, medications, and personal hygiene products. Benzoic acid dissociates in water: C6H5COOH + H2O ==== H+ and C6H5COO- The pKa of this reaction is 4.2. In a 0.505 M solution of benzoic acid, what is the fraction of the total acid that is ionized? what is the % ionized= ?

Dr.A replied: "Ka = 10^-4.2 =6.3 x 10^-5 M 6.3 x 10^-5 = x^2 / 0.505-x x = 0.00564 0.00564 x 100 / 0.505 = 1.1 %"

Please help me to calculate pH!? My question is: I want to do a buffer with benzoic acid and NaOH, and I will call this buffer when its finish for “Solution A”. I want to have 200 ml of the buffer, and I have 1,32 g benzoic acid and 24 ml of 1,00 M NaOH. C6H5COOH + OH-  C6H5COOH + H2O 1. first i want to calculate the buffer pH, and i have done like this: Vtot= 200*10-3ml mbenzoic=1,32 g Mwbenzoic=122g/mol nbenzoic= 0,01 mol VNaOH= 24*10-3 ml CNaOH= 1 M nNaOH= 0,024 mol Then I calculated the concentration in 200 ml: CNaOH= n/V => 0,024/200*10-3 = 0,12 M Cbenzoic= n/V => 0,01/200*10-3= 0,05 M Now I will use Henderson–Hasselbalch equation.. And the pKa for benzoic acid is 4,19.. So with the Henderson–Hasselbalch equation i will get the pH at the buffer: 6,59. Have i thinking right now? 2. My other question is that i want to calculate pH if I take 20 ml “Solution A” (the solution above) and establish 1ml, 1M NaOH. How do I calculate this pH? Is it the concentration for benzoic acid that I use for the calculation?

Dr.A replied: "moles acid = 1.32 g/ 122 g/mol= 0.0108 moles NaOH = 0.024 C6H5COOH + OH- => C6H5COO- + H2O all the acetic acid has been used up moles OH- in excess = 0.024 - 0.0108=0.0132 [OH-]= 0.0132 / 0.200 L= 0.066 M pOH = - log 0.066 =1.18 pH = 14 - 1.18 =12.82 Moles NaOH added = 1 x 10^-3 L x 1 M = 1 x 10^-3 moles OH- in excess = 0.0132 + 1 x 10^-3 =0.0142 total volume = 0.201 L [OH-]= 0.0142/ 0.201 =0.0706 M pOH = 1.15 pH = 12.85"

Balancing Equations Help & Question!? Could somebody please balance this equation, & possibly help with the general question? (equation to balance:) C6H5COOH-->H2O + CO2. The Actual question is : When benzoic acid is burned, the enthalpy of combustion is -3323.6 kJ/mol. Use the info & tabulated values of the standard enthalpies of formation of water (liquid) and carbon dioxide to claculate the standard enthalpy of formation of benzioc acid. ( The enthalpy of water = -285.8kJ/mol, and the standard enthalpy of carbon dioxide is = -393.5kJ/mol) Thanks a bunch, If you don't understand or know how to do the question, please at least help me balance the equation. Thankyou : )

squanto replied: "Don't forget, you don't have to use a whole number to balance an equation. Often times, water needs to be a half mole when CO2 is a product."

A sample of C6H5COOH weighing 1.221g was placed in a bomb calorimeter. ? This is ignited in a pure O2 atmosphere. A temperature rise from 25.24C to 31.67C was noted. The heat capacity of the calorimeter was 5.020kJ/C and the combustion products were CO2 and H2O. Calculate the change in H in kJ/mol for the reaction. If someone answers this, can you please show your work because I keep getting -3940kJ/mol when the answer is -3230kJ/mol. Thanks

jpsazii replied: "-3229 kJ/mol First of all, you have 1.221g C6H5COOH (benzoic acid). Since each carbon has a mass of 12.01, each oxygen has a mass of 16.00, and each hydrogen has a mass of 1.01, this compound has a molar mass of 122.13 g/mol. That means that 1.221 g is 1.221/122.13 = 0.009998 mol. Now take the change in temperature, 31.67-25.24 = 6.43. Since the heat capacity of the calorimeter is 5.020 kJ/C, the change in enthalpy can be found by taking the product of the change in temperature and the heat capacity. Since heat is given off, we give this a - sign. -(6.43 C)(5.020 kJ/C) = -32.2786 kJ. OK, so we know the change in enthalpy and the number of moles, so take the ratio. -32.2786 kJ/0.09998 mol = -3229 kJ/mol (to 4 significant figures) I hope this helps!"

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Help with an equilibrium constant problem involving acids?
The equilibrium constant for the reaction C6H5COOH(aq) + CH3COO(aq) <-> C6H5COO(aq) + CH3COOH(aq) is 3.6 at 25�C. If ... C6H5COOH: 1) C6H5COOH + H2O <==> H3O ...

balanced equation: extracted C6H5COOH (benzoic acid) from H2O ...
balanced equation: extracted C6H5COOH (benzoic acid) from H2O using CH2Cl2 ... C6H5COOH + H2O + CH2Cl2 + NaOH ... They are C6H5COOH, H2O, NaOH, and CH2Cl2. ...

Chemical Equations Balanced on 11/16/09 - Chemistry Online ...
C6H5COOH + H2O = H30 + CHCO => 60 C6H5COOH + 90 H2O = 11 H30 + 210 ... C6H5COOH + H2O = H3O + CHCO => 2 C6H5COOH + 14 H2O ... BaCl2 + CrCl3 + KCl4 + H2O + I2 ...

Can pH be solved when only Ka (acid ionization constant) and ...
C6H5COOH + H2O = H3O+(aq) + C6H5COO - (aq) Then you write out how much of each ... have to before any reaction takes place (ignoring water) C6H5COOH + H2O = H3O ...

Balance Chemical Equation - Online Balancer - Chemistry ...
2 C6H5COOH + 15 O2 = 14 CO2 + 6 H2O. Please tell about this free online software to your friends! ... C6H5COOH + O2 = CO2 + H2O ...

Organic Lab help on pH questions - Free Chemistry Help Forum ...
C6H5COO- + H3O+ C6H5COOH + H2O ... C6H5COOH + NaOH (aq) � C6H5COO- Na+ (aq) + H2O. NaOH Stronger base. C6H5COOH � Stronger acid ...

2002 Problem set 4
C6H5OH + H2O --> H3O+(aq) + C6H5O-(aq) ... C5H11N + H2O --> OH- + C5H11NH ... The equation is C6H5COOH + H2O --> C6H5COO- + H3O+ Now construct the table: C6H5COOH ...

Solution B
... mL of 0.100M benzoic acid, C6H5COOH, with 0.100M sodium hydroxide. ... C6H5COOH + H2O C6H5COO- + H3O+ (4 points) ... C6H5COO- + H2O C6H5COOH + OH- (4 points) ...

Chemistry : Chapter 5 : Chemical Reactions
... Ni(OH)22 H2O + NiS. Sr(OH)2 + C6H5COOHH2O + Sr(C6H5COO)2. Solution: It ... C6H5COOH is ... needed, add 2 OH? to that side and one H2O to the other side. ...

chapter 17. ion equilibrium: acid and base, solubility product
HCl + H2O H3O+ + Cl.? Hydration of carbon dioxide (CO2) gives carbonic acid (H2CO3) ... benzoic acid C6H5COOH + H2O H3O+ + C6H5COO? 6.3 � 10?5 4.20 ...