How many grams of H2 are needed to produce 12.88g of NH3?
scroll down to my reply to your earlier question, same method.
To determine the number of grams of H2 needed to produce 12.88g of NH3, we need to apply stoichiometry and the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction is:
N2 + 3H2 → 2NH3
This equation shows that 3 moles of hydrogen gas (H2) react with 1 mole of ammonia (NH3) to produce 2 moles of ammonia.
To find the number of grams of H2, we need to follow these steps:
Step 1: Convert the given mass of NH3 to moles.
To do this, we use the molar mass of NH3, which is 17.03 g/mol.
So, 12.88g of NH3 is:
12.88g NH3 × (1 mole NH3/17.03g NH3) = 0.7558 moles NH3
Step 2: Use the stoichiometry of the balanced equation to calculate the number of moles of H2 required.
From the balanced equation, we can see that for every 2 moles of NH3, we need 3 moles of H2.
Therefore, the ratio of moles of H2 to moles of NH3 is 3/2.
0.7558 moles NH3 × (3 moles H2/2 moles NH3) = 1.1337 moles H2
Step 3: Convert the moles of H2 to grams.
To do this, we need to use the molar mass of H2, which is 2.02 g/mol.
1.1337 moles H2 × (2.02 g H2/1 mole H2) = 2.29 grams of H2
Therefore, 2.29 grams of H2 are needed to produce 12.88 grams of NH3.