Question

4. Nitrogen gas and hydrogen gas are combined toform ammonia. What mass of ammonia can beproduced from a 10.00 gram sample of nitrogenis combined with an excess of oxygen.(N,+H2 →NH3)

Answer 1

Answer: 12.16 g of ammonia could be obtained from 10.00 g of nitrogen gas

Step-by-step explanation:

The question requires us to calculate the mass of ammonia (NH3) that could be obtained from 10.00g of nitrogen gas (N2) and excess of hydrogen (H2).

We'll consider the following unbalanced chemical equation for this reaction:

`N_2+H_2\rightarrow NH_3`

To solve this question, we need to balance the chemical equation, then calculate the amount of moles of nitrogen gas that were used and use the stoichiometry to calculate the amount of NH3 that was obtained.

We can balance the chemical equation as it follows:

- First, let's correct the amount of N atoms: there is 1 N atom on the right side and 2 on the left side. We can adjust the coefficient of NH3 from 1 to 2 to adjust this number:

`N_2+H_2\rightarrow2NH_3`

- Now, there are H atoms on the left side and 6 H atoms on the right side. We can adjust the coefficient of H2 from 1 to 3 to adjust this number. The following equation corresponds to the balanced chemical equation:

Now that we have the balanced chemical equation, we must calculate the amount of N2 moles that were used.

Knowing that the molar mass of N2 is 28.02 g/mol, we can write:

28.02 g N2 ------------------------ 1 mol N2

10.00 g N2 ------------------------- x

Solving for x, we'll have:

`x=\frac{(10.00g\text{ }N_2)*(1mol\text{ }N_2)}{(28.02g\text{ }N_2)}=0.3569mol\text{ }N_2`

Therefore, 0.3569 moles of N2 were used in the reaction.

Considering the balanced chemical equation, we can see that 1 mol of N2 is necessary to obtain 2 moles of NH3, thus we can write:

1 mol N2 ------------------- 2 mol NH3

0.3569 mol N2 --------- y

Solving for y, we'll have:

`y=\frac{(0.3569mol\text{ }N_2)*(2mol\text{ N}H_3)}{(1mol\text{ }N_2)}=0.7138mol\text{ N}H_3`

Therefore, 0.7138 moles of NH3 could be obtained.

At last, we must convert the number of moles of NH3 to its correspondent mass. Knowing that the molar mass of NH3 is 17.03 g/mol, we can write:

1 mol NH3 -------------------- 17.03 g NH3

0.7138 mol NH3 ------------ z

Solving for z, we'll have that 0.7138 moles of NH3 corresponds to 12.16 g of NH3.

Therefore, 12.16 g of ammonia could be obtained from 10.00 g of nitrogen gas.