Question

1. How many molecules of ammonia can be created when four molecules of nitrogen are combined with four molecules of hydrogen? Include a drawing of the molecules in your answer.
2. What ratio of nitrogen and hydrogen molecules would result in no left-over reactants? Explain your answer


3. If 100.0g of nitrogen is reacted with 100.0g of hydrogen, what is the theoretical yield of the reaction? What is the excess reactant? What is the limiting reactant?


Please help me I'm so confused

Answer 1

Explanation:

1. $N_2+3H_2\rightarrow 2NH_3$

According to reaction , 1 molecule of nitrogen gas reacts with 3 molecules of hydrogen gas to give 2 molecules of ammonia

Then 4 molecules of nitrogen gas will react with:

$\frac{3}{1}\times 4 molecules=12 molecules$ of hydrogen gas.

As we can see that molecules of nitrogen gas are in excess amount.

So, the molecules  of ammonia  formed will depend upon molecules of hydrogen gas.

According to reaction, 3 molecules of hydrogen gas gives 2 molecules of ammonia.

Then 4 molecules of hydrogen gas will give :

= $\frac{2}{3}\times 4 molecules=2.667 molecules$ of ammonia

2) $N_2+3H_2\rightarrow 2NH_3$

Molecules of nitrogen gas in balanced chemical equation = 1

Molecules of hydrogen gas in balanced chemical equation = 3

The ratio of nitrogen and hydrogen molecules would result in no left-over reactants will be:

$\frac{1}{3}$ = 1 : 3

For every 1 molecule of nitrogen gas molecules 3 molecules of hydrogen gas molecules are required to form.

3) $N_2+3H_2\rightarrow 2NH_3$

Moles of nitrogen gas = $\frac{100.0 g}{28 g/mol}=3.5714 mol$

Moles of hydrogen gas = $\frac{100.0 g}{2 g/mol}=50.0 mol$

According to reaction , 1 mole of nitrogen gas reacts with 3 mole of hydrogen gas to give 2 moles of ammonia

Then 3.5714 moles of nitrogen gas will react with:

$\frac{3}{1}\times 3.5714 mol=10.7142 mol$ of hydrogen gas.

As we can see that moles of nitrogen gas are only reacting with 10.7142 moles of hydrogen gas which means that nitrogen gas is limiting reagent and hydrogen gas is excessive reagent.

So, amount of ammonia gas will depend upon the moles of nitrogen gas.

According to reaction, 1 molecules of nitrogen gas gives 2 molecules of ammonia.

Then 3.5714 mol of nitrogen gas will give :

= $\frac{2}{1}\times 3.5714 mol=7.1428 mol$ of ammonia

Theoretical yield = Mass of 7.1428 moles of ammonia:

7.1428 mol × 17 g/mol = 199.9 g

199.9 g is the theoretical yield of the reaction.

Hydrogen gas  is the excess reactant.

Nitrogen gas is the limiting reactant.

Answer 2

Answer:  1. 8/3 molecules of ammonia will be produced with excess nitrogen left unreacted.

4N2(excess) + 4H2 ---------> 8/3(NH3)

2. 4/3 moles of Nitrogen: 4moles of hydrogen

3. The theoretical yield = 66.7g of NH3. The excess  reactant  is nitrogen. The limiting reactant is Hydrogen.

Explanation:

1. The balanced chemical equation for the reaction between nitrogen and hydrogen to produce ammonia is : N2 + 3H2 ------> 2(NH3).

From the above equation, it can be seen that the ratio of ammonia produced to hydrogen reacted is 2:3. So, the number of ammonia molecules produced = 4*2/3 = 8/3 moles.

2. From the balanced chemical equation of the reaction between nitrogen and hydrogen to produce ammonia above, the ratio of nitrogen to hydrogen in the reaction is 1:3. Therefore, the ratio required for a complete reaction will be 4*1/3 = 4/3 moles.

3.  The balanced chemical equation for the reaction between nitrogen and hydrogen to produce ammonia is : N2 + 3H2 ------> 2(NH3).

Therefore, 100g of hydrogen will produce 100*2/3g of ammonia = 66.7g of ammonia.

Also, 100g of hydrogen will react with 100*1/3g of nitrogen= 33.3g of nitrogen. Therefore, the limiting reactant is hydrogen and the excess reactant is nitrogen because once 100g of hydrogen is used up, the reaction stops, with about 66.7g of nitrogen left unreacted.