Zinc metal reacts with hydrochloric acid to produce zinc (II) chloride and hydrogen gas. How many liters of hydrogen gas will be produced at STP when 10.00 g of zinc metal is added to 23.8 mL of a 0.45 M hydrochloric acid solution?

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