Question

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. What is the maximum mass of H, O that can be produced by combining 57.1 g of each reactant? 4 NH3(g) + 502(g) → 4NO(g) + 6H2O(g) mass: - mass: ho

Answer 1

NH3 moles = mass / molar mass of NH3 = ( 57.1g / 17g/mol) = 3.36 mol

O2 moles =(57.1g / 32g/mol) = 1.784

we need more moles of O2 than NH3 ( as per reaction coefficients). But we have O2 is less than NH3 moles. Hence O2 is limiting reagent

H2O moles produced = ( 6/5) x O2 moles ( as per reaction coefficients )

= ( 6/5) x 1.784 = 2.14 mol

H2O mass = moles x molar mass of H2O = 2.14 mol x 18.015 g/mol

= 38.6 g