Question

4. Consider the following chemical reaction, Na(s) + 3 Hale) - 2 NH,(g) [balanced] 15:42 g of nitrogen gas are reacted with 5.42 g of hydrogen gas, which of the reactants is the limiting reactant? Use the molar mass data below if necessary. Show all your work to explain your answer. (24 points) Molar mass of N7 - 28.02 g/mol Molar mass of H; -2.02 g/mol Molar mass of NH) - 17.04 g/mol | N2 + 3H₂ → 2NH3 | 28g leg 349 15,42 5.42 289.N2 - legt 15.42 g N2 = 4x5,42 1 28 = 1.16H₂ We only need one lilleg of H2 but we have 5,42 g. The linuting renetant would be N2

Answer 1

Given molar mass of N₂ is 28.02 g/mol

Molar mass of H₂ is 2.02 g/mol

According to the stoichiometric equation 1mole of N₂ reacts with 3moles of H₂ to give 1mole of NH₃

This means 1x28.02 gms of N₂ reacts with 3x2.02 gms of H₂

Given 5.42 gms of N₂ and 5.42 gms of H₂ are allowed to react

If 6.06 gms of H₂ can react with 28.02 gms of N₂ then

5.42 gms of H₂ can react with

Mass of Nitrogen can be consumed = (5.42/6.06)×28.02 = 25.11 gms

Thus 6.06 gms of hydrogen can react with 25.11 gms of nitrogen. But the available nitrogen is only 5.42gms. As the availability of nitrogen become limited and still hydrogen is present, here nitrogen is the limiting reagent.