Question

1:How many moles of ammonia will be produced if 12 moles of nitrogen are used?

2:How many moles of nitrogen are required to react with 6 moles of hydrogen?

3:Honors: If there are 40 moles of nitrogen and 30 moles of hydrogen, what is the limiting reactant? Explain your answer.

4:Honors: What is the mass of hydrogen required to produce 80 g of ammonia?

Please, I really need help, so I can graduate

Answer 1

See Explanation

Step-by-step explanation:

The reaction equation is

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

1)If 1 mole of N2 yields 2 moles of NH3

12 moles of N2 will yield 12 * 2/1 = 24 moles of NH3

2) 3 moles of nitrogen reacts with 1 mole of nitrogen

6 moles of hydrogen will react with 6 * 1/3 = 2 moles of nitrogen

3) The limiting reactant yields the least amount of product

If 1 mole of nitrogen yields 2 moles of NH3

40 moles of nitrogen yields 40 * 2/1 = 80 moles of NH3

If 3 moles of hydrogen yields 2 moles of NH3

30 moles of hydrogen yields 30 * 2/3 = 20 moles of NH3

Hence hydrogen is the limiting reactant

4) Number of moles in 80g of NH3 = 80g/17g/mol = 4.71 moles

3 moles of hydrogen produces 2 moles of ammonia

x moles of hydrogen produces 4.71 moles of ammonia

x = 3 * 4.71/2 = 7.065 moles of hydrogen

Mass = 7.065 moles of hydrogen * 2 g/mol

Mass = 14.13 g of hydrogen