Need help with this question :)

Question

A chemical reaction produces 456 mL of nitrogen dioxide (NO₂) gas at STP. What is the mass of the gass produced?

Answer 1

Since the system is at STP, you don't have to use the ideal gas law (although there's nothing wrong with doing it that way). To save time, you can use the fact that at STP, 1 mol of gas = 22.4 L.

456 ml = 0.456 L

0.456 L x 1 mol/22.4 L x 46 g NO₂/mol NO₂ = 0.936 g NO₂

Answer 2

When working with a sample of gas, remember that we can relate pressure, volume, absolute temperature, and the number of moles of gas using the Ideal Gas Law (PV = nRT). In addition, if we know the number of moles of our gas, we can find the number of grams using the molar mass. Thus, we'll start by finding the number of moles of NO₂ from the volume of NO₂, then convert that number of moles of NO₂ into grams of NO₂.

Note: STP is defined as 1 atm and 0^oC (or 273 K). We also have the value of the gas constant R = 0.0821 L*atm/K*mol

PV = nRT; (1 atm)(0.456 L) = (n)(0.0821 L*atm/K*mol)(273 K); n = 0.0203 mol NO₂

0.0203 mol NO₂ * 46.005 g NO₂/1 mol NO₂ = 0.934 g NO₂

I hope this helped!