How many moles of H_2 are needed to produce 13.27g of NH_3?
well one mole of NH3 has 3 moles of H
1 mole of H2 has 2 moles of H
each mole of NH3 requires 3/2 moles of H2
mol wt of NH3 = 14 + 3*1 = 17
so you need 3/2 * (13.27 / 17) = 1.17
To find how many moles of H2 are needed to produce 13.27g of NH3, we need to use the balanced chemical equation for the reaction. The balanced chemical equation for the reaction between H2 and NH3 is:
3H2 + N2 -> 2NH3
From the balanced equation, we can see that it takes 3 moles of H2 to produce 2 moles of NH3.
To find the number of moles of H2 needed, we can use the following calculation:
moles of H2 = (mass of NH3 * moles of H2 in the balanced equation) / molar mass of NH3
First, calculate the molar mass of NH3:
N has a molar mass of 14.01 g/mol
H has a molar mass of 1.01 g/mol (there are 3 H atoms in NH3)
So, the molar mass of NH3 is:
(14.01 g/mol) + (3 * 1.01 g/mol) = 17.04 g/mol
Now, plug in the values into the formula:
moles of H2 = (13.27g * 3/2) / 17.04 g/mol
Calculate the moles of H2:
(13.27g * 3/2) / 17.04 g/mol ≈ 2.47 moles
Therefore, approximately 2.47 moles of H2 are needed to produce 13.27g of NH3.