The Grass Doctor

Chem question I need help with?

1. Urea, (NH2)2CO, is an important fertilizer that is manufactured by the following reaction: 2NH3 + CO2 --> (NH2)2CO + H2O What volume of NH3 at STP will be needed to produce 7.50 x 10^4 kg of urea if there is an 89.5% yield in the process?

Public Comments

1. First calculate what is 89.5% of your given amount of urea. 7.5 x 10^4 kg * .895 = 67,125g urea 67,125g urea * (1 mol urea / 60g urea) * (2 mol NH3 / 1 mol urea) = 2237.5 mol NH3 (thats a lot...) PV = nRT 1 atm * V = 2237.5mol * .08205 L * atm / mol * k * 273K V = 50119L