Step-by-step explanation:
First, we need to write the equation balanced:
N2 + 3 H2 --> 2 NH3
If 1 batch contains 175 grams:
1 batch --- 175 grams
70 batches ---- x
x = 12,250 g of ammonia needs to be produced.
Now let's transform 22,000 g of N2 and 1,000 grams of H2 into moles. For this, we use the following formula:
n = m/MM
MM of N2: 28 g/mol
n = 22,000/28
n = 785.7 moles
MM of H2: 2 g/mol
n = 1000/2
n = 500 moles
So it means that H2 is the limiting reactant, because the stoichometry of N2 and H2 is 1:3.
So:
3 moles of H2 --- 2 moles of NH3
500 moles of H2 --- x moles of NH3
x = 333.33 moles of NH3
Now we transform this value into grams:
m = n*MM
MM of NH3 = 17 g/mol
m = 333.33*17
m = 5,666.6 g
With 22,000 g of N2 and 1000 g of H2 it will be produced 5,666.6 g of NH3.
We need to produce 12,250 g of ammonia.
So:
Answer: B. No, the hydrogen will run out first.