Question

STOICHIOMETERY QUESTION :-When aluminum metal is added to sulphuric acid, hydrogen gas and aluminum sulphate is produced. What volume of hydrogen gas will evolve if we use 15.00 mL of a 0.250M solution of sulphuric acid and add 0.0712 g of aluminum? Assume that this reaction is only 90.0% efficient and that the experiment was performed at 1.00 atm and 25.0°C. (MMAl = 26.98 g/mole)

Answer 1

V of H2? if

V = 15 ml of H2SO4

M = 0.25 M of H2SO4

m = 0.0712 g of Al

x = 90%

P = 1 atm and T = 25 C or 298K

MW Al = 26.98 g/mol

First, the reaction

2Al + 3H2SO4 --> 3H2 + Al2(SO4)3

find moles of acid used

mol =M*V= 15/1000 L * 0.25 M = 0.00375 mol of Acid used

find moles of Aluminium present

mol = m/MW = 0.0712 g/ 26.98 g/mol = 0.00263 mol of Aluminium

3 mol of Acid need 2 mol of Aluminium, then, find limiting reactant

0.00375 mol of Acid would need 2/3 of Aluminium , that is 0.00375*2/3 = 0.0025 mol of Aluminiumn neded (which we have... in excess, wo the acid is the limiting reactant)

Now,

let us check how much H2 we are going to produce

3 mol of Acid will produce 3 mol of H2 gas, therefore

0.00375 mol of Acid produce 0.00375 mol of H2

but this process is only 90% effective so

0.90*0.00375 = 0.003375 mol of H2 will be produced

We can find out the volume using ideal gas law

PV= nRT

V = nRT/P

R = 0.082

V = 0.003375 *0.082 * (298) / 1atm = 0.08247 L

V = 82.47 ml of H2