meA)What is the molarity of C6H12O6 in an aqueous solution that is prepared by d

Question

meA)What is the molarity of C6H12O6 in an aqueous solution that is prepared by dissolving a 20.95 g sample of impure C6H12O6 in water to make 1,450 mL of solution. Assume that the sample is 93% C6H12O6 by mass.Enter your answer in mol/L, accurate to three significant figures
B)An aqueous stock solution is 28.0% NH3 by mass and its density is 0.9 g/mL. What volume of this solution is required to make 1.50 L of 2.75 mol/L NH3(aq)? Give your answer in millilitres, accurate to three significant figures.
C)What is the minimum volume of 2.85 mol/L HCl(aq) required to dissolve 12.1 g Zn metal? The atomic weight of Zn is 65.39 g/mol. The unbalanced chemical equation for the reaction is
Zn(s) + HCl(aq) --> ZnCl2(aq) + H2(g)Enter your answer accurate to three significant figures
D)What volume of 2.09 mol/L H2SO4(aq) is required to completely neutralize 40.5 mL of 1.47 mol/L KOH(aq) ?Give your answer in mL, accurate to three significant figures
E)A 5.645-g sample of a mixture of Fe and Al is treated with excess HCl(aq). If 0.2122 moles of H2 are obtained, then what is the percentage by mass of Fe in the original mixture? Give your answer accurate to three significant figures

Explanation / Answer

A) Mass of C6H12O6 = 93*20.95/100 = 19.4835 g

Volume of solution = 1450 mL = 1.45 L

Moles of C6H12O6 = 19.4835/(72+12+96) = 0.1082 moles

Concentration = 0.1082/1.45 = 0.0746 0.075 mol/L

B) Moles of NH3 Required = 1.5*2.75 = 4.125 moles

Let the volume required is x ml

Then mass of solution = 0.9x g

Mass of NH3 = 28*0.9x/100

Moles of NH3 = 28*0.9x/(100*17) = 4.125

=> x = 278.27 278 mL

C) Balanced Chemical equation

Zn (s) + 2 HCl (aq) -----> ZnCl2 (aq) + H2 (g)

SO 2 moles of HCL is required for 1 mole of Zn

moles of Zn = 12.1/65.39 = 0.185

moles of HCL requires = 2*0.185 = V*2.85

=> V = 0.1298 0.130 L

D) 2KOH + H2SO4 -------> 2K2SO4 + 2 H2O

Let the volume required = V

2*V*2.09 = 40.5*1.47 => V = 14.243 14.2 mL

E) Sue to excess HCl Fe will also be oxidised to +2 state

Fe + 2 HCl ---------> FeCl2 + H2

2 Al + 6 HCl ---------> 2AlCl3 + 3H2

Let there are x moles of Fe & y moles Of Al

moles of H2 produced will be = 2x + 3 y = 0.2122 ------- eqn 1

Equalling weight

56x + 27y = 5.645 ------------- eqn 2

solving the equations we get

x = 0.0983 & y = 0.0052

Mass of Fe = 56x = 5.5048 g

Mass Of Al = 27y = 0.1404 g

% ratio of Fe = 5.5408*100/(5.5408+0.1404) = 97.53 % 97.5 %

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