Determine the number of moles of air present in 1.35 l at 760 torr and 17.0 c. which equation should you use?

C. N = PV/RT is the anwser to your question

Generally PV=nRT so see attached.

Explanation:

244.7 g of acetylene

Explanation:

The balanced reaction equation is shown below;

2H20 (l) + CaC2 (s) → Ca(OH)2 (s) + C2H2 (g)

Number of moles of were reacted = reacting mass/molar mass = 358g/18gmol-1 = 19.89 moles of water

From the balanced reaction equation;

2 moles water yields 1 mole of acetylene

19.89 moles of water will yield 19.89 × 1/2 = 9.945 moles of acetylene

Theoretical yield of acetylene = 9.945 moles of acetylene × molar mass of acetylene

Molar mass of acetylene = 26.04 g/mol

Theoretical yield of acetylene = 9.945 moles of acetylene × 26.04 g/mol

Theoretical yield of acetylene = 258.9678 g of acetylene

% yield = actual yield/ theoretical yield × 100

94.5 = actual yield/258.9678 g × 100

Actual yield= 94.5 × 258.9678 g/100

Actual yield = 244.7 g of acetylene