If 100 g of H2O is produced from 16 g of H2, What is the % yield of H2O?
(Molar mass: H= 1g/mol; O=16 g/mol)
H2 + O2 → 2H2O
First, let's calculate how many moles of H2 were used:
16 g H2 / 2 g/mol = 8 mol H2
According to the balanced equation, 1 mol of H2 produces 2 mol of H2O. Therefore, the theoretical yield of H2O can be calculated as:
8 mol H2 x 2 mol H2O/mol H2 = 16 mol H2O
Now, let's calculate the actual yield. We were given that 100 g of H2O was produced, which can be converted to moles:
100 g H2O / 18 g/mol = 5.56 mol H2O
The percent yield can then be calculated as:
(actual yield / theoretical yield) x 100%
= (5.56 mol H2O / 16 mol H2O) x 100%
= 34.8%
Therefore, the percent yield of H2O is 34.8%. This indicates that the reaction did not produce as much H2O as would be expected from the amount of H2 used, and some H2 may have been lost or reacted in other ways.
To calculate the percent yield of H2O, we need to compare the actual yield of H2O to the theoretical yield.
1. Calculate the theoretical yield of H2O:
Since the balanced equation shows that 2 moles of H2 are required to produce 2 moles of H2O, we can find the theoretical yield of H2O using the stoichiometry.
Molar mass of H2O = (2 * molar mass of H) + molar mass of O
= (2 * 1 g/mol) + 16 g/mol
= 18 g/mol
So, for every 2 moles of H2, we get 18 grams of H2O.
Using the given 16 grams of H2, we can calculate the theoretical yield of H2O:
Theoretical yield of H2O = (mass of H2 / molar mass of H2) * (molar mass of H2O / coefficient of H2O)
= (16 g / 2 g/mol) * (18 g/mol / 2)
= 8 * 9
= 72 g
2. Calculate the percent yield:
To find the percent yield, divide the actual yield by the theoretical yield and multiply by 100.
Actual yield of H2O = 100 g (given in the question)
Percent yield of H2O = (actual yield / theoretical yield) * 100
= (100 g / 72 g) * 100
= 138.88%
Therefore, the percent yield of H2O is approximately 138.88%.