Question

Please explain the best you can, having trouble learning this for my test in two days.

What volume of H2 (g) ( at 750. mmHg and 25 °C ) is produced from 50.0 mL of H2SO4 and 0.300 g of Al?

2Al (s) + 3H2SO4 (aq) → Al2(SO4)3 (aq) + 3 H2 (g)
Question 2 options:

0.413 L

0.265 L

0.448 L

0.222 L

0.565 L

Answer 1

STEP #1: Find the limiting reagent. You are given quantities of both reactants, so you need to convert each to moles and figure out which reactant is limiting. You can find moles of Al by dividing the mass by the molar mass (n=m/MM). You need a concentration of H2SO4 to find moles if all you're given is volume (C=n/V).

STEP #2: Use stoichiometry and the moles of limiting reagent to figure out how many moles of H2 are produced.

STEP #3: You now have moles of H2, in addition to pressure and temperature (given in the question). Using PV=nRT, you can find volume!