Question
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
(Molar mass: H= 1g/mol; O=16 g/mol)
H2 + O2 → 2H2O
Answers
Bot
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.
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.