Answer:
0.120 L of hydrogen gas will be produced
Explanation:
Step 1: Data given
Mass of zinc = 10.0 grams
Volume of hydrochloric acid = 23.8 mL
Molarity of hydrochloric acid = 0.45 M
Molar mass of zinc =65.38 g/mol
Step 2: The balanced equation
Zn + 2HCl → ZnCl2 + H2
Step 3: Calculate moles Zinc
Moles Zn = mass Zn / molar mass Zn
Moles Zn = 10.0 grams / 65.38 g/mol
Moles Zn = 0.153 moles
Step 4: Calculate moles HCl
Moles HCl = molarity * volume
Moles HCl = 0.45 M * 0.0238 L
Moles HCl = 0.01071 moles
Step 5: Calculate limiting reactant
For 1 mol Zn, we need 2 moles HCl to produce 1 mol ZnCl2 and 1 mol H2
HCl is the limiting reactant. It will completely be consumed (0.01071 moles)
Zn is in excess. There will react 0.01071/2 = 0.005355 moles
There will remain 0.153 - 0.005355 = 0.147645 moles
Step 6: Calculate moles H2
For 1 mol Zn, we need 2 moles HCl to produce 1 mol ZnCl2 and 1 mol H2
For 0.01071 moles HCl we'll have 0.005355 moles H2
Step 7: Calculate volume H2
1 mol at STP = 22.4 L
0.005355 moles = 22.4 * 0.005355 = 0.120 L = 120 mL
0.120 L of hydrogen gas will be produced
