# Question #7fb80

##### 1 Answer

You'd need **52.2 g** of sodium azide to produce that much nitrogen gas under those conditions.

Start with the balanced chemical equation

Notice that you ahve a **3/2 times** more nitrogen gas.

Since you already know *what your reaction must produce*, you can works backwards to determine how much sodium azide must react.

Since you know volume, temperature, and pressure, you can use the ideal gas law equation to determine how many moles of nitrogen gas your reaction must produce

According to the aforementioned mole ratio, this many moles of nitrigen would require

Now use sodium azide's molar mass to determine how many grams you'd need