2NaN3(s) −> 2Na(s) + 3N2(g) .
If 0.65 mol of NaN3 reacts, what mass of
nitrogen would result?
Answer in units of g.
Step by step with answer please!
I'm begging!
Certainly! Let's break down the steps to find the mass of nitrogen resulting from the reaction.
1. Start with the given amount in moles: 0.65 mol of NaN3.
2. Determine the molar ratio between NaN3 and N2 from the balanced equation:
According to the balanced equation, 2 moles of NaN3 produce 3 moles of N2.
3. Use the molar ratio to convert moles of NaN3 to moles of N2:
(0.65 mol NaN3) x (3 mol N2 / 2 mol NaN3) = 0.975 mol N2
4. Convert moles of N2 to grams using the molar mass of nitrogen:
The molar mass of nitrogen (N2) is approximately 28.02 g/mol.
(0.975 mol N2) x (28.02 g N2 / 1 mol N2) = 27.26 g N2
Therefore, the mass of nitrogen resulting from the reaction of 0.65 mol of NaN3 is 27.26 grams.