# 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?

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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

#### 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