How many grams of H2 are needed to produce 12.22g of NH3?
IF you are using the equation of
N2 + 3H2 ==> 2NH3 (if not this won't work)
mols NH3 = grams/molar mass
mols H2 = mols NH3 x (3/2) = ?
Then g H2 = mols H2 x molar mass H2 = ?
To determine the grams of H2 needed to produce 12.22g of NH3 (ammonia), we need to follow a balanced chemical equation that relates the two.
The balanced chemical equation for the reaction between H2 and NH3 is:
3H2 + N2 → 2NH3
From the equation, we can see that 3 moles of H2 react to produce 2 moles of NH3. The molar mass of H2 is about 2 g/mol, and the molar mass of NH3 is about 17 g/mol.
To find the number of moles of NH3 produced from 12.22g, we can use the equation:
moles = mass / molar mass
moles of NH3 = 12.22g / 17 g/mol
≈ 0.719 moles
Now, using the ratio of moles from the balanced equation, we can find the moles of H2 needed to produce the calculated moles of NH3:
(3 moles H2 / 2 moles NH3) × 0.719 moles NH3
≈ 1.079 moles H2
Finally, we can convert the moles of H2 to grams by multiplying by the molar mass of H2:
grams of H2 = 1.079 moles H2 × 2 g/mol
≈ 2.158 grams of H2
Therefore, approximately 2.158 grams of H2 are needed to produce 12.22 grams of NH3.