# At #"1.20 atm"#, and a temperature of #"27"^@"C"#, #"75.0 L NH"_3"# gas is produced by the reaction between #"N"_2"# gas and #"H"_2"# gas. How many moles and what volume of #"N"_2"# and #"H"_2"# are required to produce #"75.0 L NH"_3"#?

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The reaction is #"N"_2("g") + "3H"_2("g")"# #rarr# #"2NH"_3("g")"# .

The reaction is

##### 2 Answers

#### Answer:

The moles of

The moles and volume of

The moles and volume of

Answers are rounded to three significant figures.

#### Explanation:

**Balanced Equation**

Use the ideal gas law with the equation

**Known/Given:**

**All Gases:**

**Ammonia #"NH"_3"#**

**Unknowns:**

**Determine the moles Ammonia #"NH"_3"#.**

**Determine the moles and volume of #"N"_2"#.**

Multiply the moles

Determine the volume of

Rearrange the ideal gas law equation to isolate

**Determine the moles and volume of #"H"_2"# at #"1.20 atm"# and #"300 K"#.**

Multiply the moles

Determine the volume of

Rearrange the ideal gas law equation to isolate

#### Answer:

Here's an alternative way of solving this problem.

#### Explanation:

A very important thing to notice here is that you're dealing with gases that are **at the same pressure and temperature**. This means that their mole ratios become equivalent to their **volume ratio**.

So, for any two gases that take part in this reaction, you have

This means that you can determine the volume of nitrogen and hydrogen by

and

So, once you use the ideal gas law equation to get the number of moles of ammonia, all you have to do is use the mole ratios.

Rounded to three sig figs, you have

Therefore, you get

and