4. How many grams of acetylene would be given off if the limiting reagent in the reaction is calcium carbide and you start with 2.00 g

0.78 g of acetylene.

Explanation:

The equation of the reaction is;

CaC2(s) + 2H2O (l) > Ca(OH)2(aq) + C2H2(g)

We have been told that CaC2 is the limiting reactant.

Number of moles of CaC2 = 2g/ 64 g/mol = 0.03 mol

If 1 mole of CaC2 yields 1 mole of acetylene

0.03 moles of CaC2 yields 0.03 moles of acetylene

Mass of acetylene = number of moles * molar mass

Molar mass of acetylene = 26 g/mol

Mass of acetylene =  0.03 moles * 26 g/mol

Mass of acetylene = 0.78 g of acetylene.