Question

For the following reaction, 8.29 grams of nitrogen gas are allowed to react with 1.05 grams of hydrogen gas . nitrogen(g) + hydrogen(g) — ammonia(g) What is the maximum mass of ammonia that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams

Answer 1

Answer

maximum anount of NH3 that can be formed: 5.905g

formula of the limiting reagent: H₂

mass excess remains after the reaction is complete: 3.435g

Explanation

N2(g) + 3H2(g) ------> 2NH3(g)

Stoichiometrically, 1mole of N2 react with 3moles of H2

given moles of N2 = 8.29g/28.02g/mol = 0.29586mol

given moles of H2 = 1.05g / 2.02g/mol = 0.51980mol

0.29586mol of N2 require 0.885758moles of H2 but available moles of H2 is 0.51980mol ,so

limiting reagent is H2

stoichiometrically , 3 moles of H2 gives 2moles of NH3

number of moles of NH3 obtained by 0.51980moles of H2 = (2/3)× 0.51980mol = 0.34653mol

mass of NH3 formed = 0.34653mol × 17.04g/mol = 5.905g

0.51980moles of H2 react with 0.17327moles of N2

remaining moles of N2 after completion of reaction = 0.29586 mol - 0.17327mol = 0.12259mol

mass of N2 remaining after completion of reaction = 0.12259mol × 28.02g/mol = 3.435g