Answer to Question #83456 in General Chemistry for stephanie

Question

For the following reaction,
3 Na(s) + P(g)  Na3P(s)
50.00 g of sodium and 60.00 L of phosphorus gas at 755 torr and 645°C were combined. How many grams of sodium phosphide were produced?

Answer 1

Answer:

Balanced reaction is 3Na(s) + P(g) = Na3P(s) first calculte mole of P gas Use ideal gas equation for calculation of mole of gas Ideal gas equation PV = nRT where, P = atm pressure= 755 torr = 0.993421 atm, V = volume in Liter = 60 L n = number of mole, T = temperature in K = 918.15 K, R = 0.0821 l*atm / mole*K

n (P) = PV/RT

n (P) = 0.993421*60 / 0.0821*918.15 = 0.79 mole

n (Na) = 50/23 = 2.17 mole

Thus, sodium is the limiting reactant and phosphorus is the excess reactant.

n (Na3P) = 0.72 mole

m (Na3P) = 0.72*99.94 = 71.95 g