What volume of nitrogen gas will be produced if 117.0 g of sodium azide is decomposed under standard conditions of 20.2°C and 101.2 kPa?

Convert the given mass of sodium azide to moles.
Use the mole ratio from the balanced equation to find the moles of nitrogen gas.
Use the ideal gas law to convert the moles of nitrogen gas to volume.

Step 1: Convert the given mass of sodium azide to moles.

The molar mass of sodium azide (NaN3) is 65.01 g/mol. To convert grams to moles, divide the given mass by the molar mass:

117.0 g / 65.01 g/mol = 1.799 mol

Step 2: Use the mole ratio from the balanced equation to find the moles of nitrogen gas.

According to the balanced equation, 2 moles of sodium azide produce 3 moles of nitrogen gas. Therefore, we can set up a ratio:

1.799 mol NaN3 / 2 mol NaN3 = x mol N2 / 3 mol N2

Solving for x, we find:

x = (1.799 mol NaN3 * 3 mol N2) / 2 mol NaN3 = 2.6985 mol N2

Step 3: Use the ideal gas law to convert the moles of nitrogen gas to volume.

The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Given that the temperature is 20.2 °C, we need to convert it to Kelvin:

T = 20.2 + 273.15 = 293.35 K

Assuming the pressure is 101.2 kPa, we can rearrange the ideal gas law to solve for volume:

V = (nRT) / P = (2.6985 mol N2 * 8.314 L·kPa/mol·K * 293.35 K) / 101.2 kPa = 63.02 L

Therefore, the volume of nitrogen gas produced is 63.02 L.

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