Question

If 4.55 mol of calcium carbide (CaC2) reacts with an excess of water, how many moles of acetylene (C2H2), a gas used in welding, will be produced?

Answer 1

There will be produced 4.55 moles C2H2 ( this is 118.48 grams)

Step-by-step explanation:

Step 1: Data given

Number of moles CaC2 = 4.55 moles

water is in excess

Molar mass of CaC2 = 64.1 g/mol

Molar mass of acetylene = 26.04 g/mol

Step 2: The balanced equation

CaC2 + 2H2O → C2H2 + Ca(OH)2

Step 3: Calculate moles of C2H2

For 1 mol CaC2 we need 2 moles of H2O to produce 1 mol of C2H2 and 1 mol of Ca(OH)2

For 4.55 moles of CaC2 we'll have 4.55 moles C2H2

Mass C2H2 = 4.55 moles C2H2 *26.04 g/mol = 118.48 grams

There will be produced 4.55 moles C2H2 ( this is 118.48 grams)

Answer 2

4.55 mol of C₂H₂

Step-by-step explanation:

Let's verify the reaction:

CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂

Ratio is 1:1, so 1 mol of calcium carbide will produce 1 mol of acetylene.

4.55 mol of CaC₂ will make the same amount of C₂H₂