Question

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`?

Answer 1

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;

`"28.01 g of"color(white)(l) N_2` needs `"6.048 g of"color(white)(l) H_2`

`"1 g of" color(white)(l)N_2` needs `6.048/28.01 "g of"color(white)(l) H_2`

`"50.0 g of"color(white)(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`

Answer 2

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 -: 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 -: M`", you need to find n.
The number of moles (n) for nitrogen is:
[`n = 50.0 -: 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 `(3H_2)`.

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 xx 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 -: 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 xx M`.

The mass (m) of hydrogen is:
[`m = 5.357142857 xx 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.