Question

Use the following equation to answer the following question

3 H2 + Na2 —-> 2 NH3

a) Determine the limiting reactant when 5.78 g of H2, and 6.28 g N2 are reacted to make NH3 you expect to get out of the chemical reaction.

Answer 1

Answer: nitrogen

Step-by-step explanation:

1. Converting 5.78 g of hydrogen to moles, we know that the formula mass of hydrogen is about 2(1.00794)=2.01588 g/mol, so 5.78 grams is about 5.78/2.01588=2.867 mol.
2. Converting 6.28 g of nitrogen to moles, we know that the formula mass of nitrogen is about 2(14.0067)=28.0134 g/mol, so 6.28 grams is about 6.28/28.0134 = 0.22417 mol.

From the equation, we know that for every 3 moles of hydrogen consumed, 1 mole of nitrogen is consumed.

• Considering the hydrogen, the reaction can occur 2.867/3=0.955 times.
• Considering the nitrogen, the reaction can occur 0.22417 times.

Therefore, nitrogen is the limiting reactant.