Question

If you need 3L of NH3 how many grams of N2 are required?3H2(g) + N2(g) ==> 2NH3(g)

Answer 1

1.88 grams

Step-by-step explanation

Given:

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

Volume of NH3(g) = 3 L

What to find:

The grams of N2(g) required.

Step-by-step solution:

From this balanced equation, 3 moles of H2(g) react with 1 mole N2(g) to produce 2 moles NH3(g).

That can also mean 3 liters of H2(g) reacts with 1 liter N2(g) to produce 2 liters of NH3(g).

Thus, if it takes 1 liter of N2(g) to produce 2 liters of NH3(g),

Then it will take x liters of N2(g) to produce 3 liters of NH3(g)

To get x, cross multiply and divide both sides by 2 liters NH3(g):

`x=\frac{1\text{ }liter\text{ }N_2(g)*3\text{ }liters\text{ }NH_3(g)}{2\text{ }liters\text{ }NH_3(g)}=1.5\text{ }liters\text{ }N_2(g)`

1.5 liters of N2(g) are required to produce 3 L of NH3(g).

The final step is to convert 1.5 liters of N2(g) to grams.

1 mole of N2(g) occupies 22.4 L at STP

Note: 1 mole of N2(g) = 28.01 g

Thus, if 28.01 grams N2(g) occupies 22.4 L at STP

Then the mass of N2(g) in 1.5 L of N2(g) is:

`\frac{1.5\text{ }L*28.01\text{ }grams}{22.4\text{ }L}=1.88\text{ }grams`

Therefore, the grams of N2(g) required to 3L of NH3(g) is 1.88 grams