Given the following, how many grams of water could be produced in this reaction? What is the limiting reactant? If 8.7 g of water are produced, what is the percent yield?
13.74 L of hydrogen gas at 30.0 C and 801 torr and 6.55 L of oxygen gas at 25 C and 801 torr are drawn into a cylinder where the following reaction takes place. #2 H_2 + O_2(g) -> H_2O (l)#
13.74 L of hydrogen gas at 30.0 C and 801 torr and 6.55 L of oxygen gas at 25 C and 801 torr are drawn into a cylinder where the following reaction takes place.
1 Answer
Explanation:
Note: this equation is not fully balanced; here is what it should look like:
To solve this equation, let's first find the limiting reactant, by using the ideal-gas equation to find the moles of each reactant present:
The units for pressure must be in
The temperature must also be in
Let's now rearrange the ideal-gas equation to solve for the number of moles,
and plug in the values for each reactant to find their quantity:
We find the limiting reactant by dividing each by the coefficient in front of it in the chemical equation; the lower value is the limiting reactant:
Thus, hydrogen is the limiting reactant, so we'll use the mole values of
To find the number of grams of
To calculate this value in grams, we'll use its molar mass (
If