# How many grams of hydrogen are necessary to react completely with 50.0 g of nitrogen in the reaction N_2+3H_2->2NH_3?

Apr 17, 2016

10.80 g of ${H}_{2}$

#### Explanation:

As per the equation, let us calculate the mole ratio.

N_2 + 3H_2 → 2NH_3

As per the equation one mole of nitrogen reacts with 1 mol of hydrogen.

In terms of mass. 28.01 g of nitrogen needs 3 mol of hydrogen or 6.048 g of hydrogen.

We can set up the ratio;

$\text{28.01 g of} \textcolor{w h i t e}{l} {N}_{2}$ needs $\text{6.048 g of} \textcolor{w h i t e}{l} {H}_{2}$

$\text{1 g of} \textcolor{w h i t e}{l} {N}_{2}$ needs $\frac{6.048}{28.01} \text{g of} \textcolor{w h i t e}{l} {H}_{2}$

$\text{50.0 g of} \textcolor{w h i t e}{l} {N}_{2}$ needs (6.048 × 50.0)/28.01color(white)(l) "g of"color(white)(l) H_2 = "10.80 g of"color(white)(l) H_2

Apr 26, 2016

See below..

#### Explanation:

Assume that:
n = number of moles
m = mass of substance
M = molar mass (equivalent to atomic weight on the periodic table)

$n = m \div M$

The mass (m) of ${N}_{2}$ has been given to you: 50.0 grams.

You first need to find out the molar mass (M) of nitrogen. If you refer to your periodic table, the molar mass (M) of nitrogen is 14.0 g/mol.
Since there are two nitrogens (refer to equation - look carefully), the molar mass of ${N}_{2}$ in this reaction is 28.0 g/mol.

Now you need to calculate the number of moles (n) for nitrogen. If you refer back to the formula "$n = m \div M$", you need to find n.
The number of moles (n) for nitrogen is:
[$n = 50.0 \div 28.0$] = 1.785714286 moles.
Note: Don't round it off just yet because this is not the final answer - or you'll get inaccurate answer.

You next step is to determine the number of moles (n) for hydrogen $\left(3 {H}_{2}\right)$.

Since the mole ratio between ${N}_{2} : {H}_{2}$ is 1 : 3, so,
If 1 mole of ${N}_{2}$ gives you 3 moles of ${H}_{2}$,
then 1.785714286 moles of ${N}_{2}$ should give you:
[$1.785714286 \times 3$] = 5.357142857 moles of ${H}_{2}$.

Now you know the number of moles for ${H}_{2}$.

The next step is to find out the molar mass (M) of Hydrogen. If you refer to your periodic table, you know that hydrogen's molar mass is 1.0 g/mol.

Since there are two hydrogens present the molar mass of hydrogen in this reaction is 2.0 g/mol.
(look at equation, carefully - there are 2, not 5 or 6 because the front number is only used for moles).

Your last step is to find out its mass (m).

If you refer back to the formula $n = m \div M$, you need to make (m) as the subject because you are finding the mass.
Once you've make m as the subject, the formula will look $m = n \times M$.

The mass (m) of hydrogen is:
[$m = 5.357142857 \times 2.0$] = 10.7 grams.
Note that: this answer is rounded to 3 significant figures.

Therefore, 10.7 grams of Hydrogen are necessary to fully react with 50.0 grams of Nitrogen.