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

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.