Aluminum metal shaving (10g) was placed in 100mL of 6.00M hydrochloric acid. Wha

Question

Aluminum metal shaving (10g) was placed in 100mL of 6.00M hydrochloric acid. What is the mass of hydrogen gas that can be produced? 2Al (s) + 6HCl (aq) -----> 2AlCl3 + 3H2 (g)

Explanation / Answer

Find the moles of Al: n = m/Ar = 10/27 = 0.37 mol (approx) where Ar = 27 is the relative atomic mass of Al. Find the moles of HCl: n = C*V = 6*0.1 = 0.6 mol HCl According to the balanced chem. equation: 2Al + 6HCl --> 2AlCl3 + 3H2 1 mol of Al reacts with 3 mol of HCl, so 0.37 mol of Al reacts with 3*0.37 = 1.11 mol of HCl So HCl is the limiting reactant and because: 6 mol of HCl produces 3 mol H2 0.6 mol of HCl produces 0.3 mol H2 This quantity - 0.3 mol of H2 - at STP occupies: V = n*Vm = 0.3*22.4 = 6.72 L, where Vm = 22.4 L/mol is the molar volume in STP.

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