Answer:
1) 65.0
2) 16.434 L = 16434 mL.
Explanation:
<em>2NaN₃ → 2Na + 3N₂,</em>
- It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.
<em>Q1: How many grams of NaN₃ are needed to make 23.6L of N₂? </em>
Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.
- Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:
1.0 L of N₂ contains → 0.92 g of N₂.
23.6 L of N₂ contains → ??? g of N₂.
∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.
- We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:
n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.
- We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:
<em><u>using cross multiplication:</u></em>
2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.
??? mol of NaN₃ produce → 0.775 moles of N₂.
∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.
- Finally, we can get the grams of NaN₃ needed:
<em>mass = no. of moles x molar mass</em> = (0.517 mol)(65.0 g/mol) =<em> 33.6 g.</em>
<em />
<em>Q2: How many mL of N₂ result if 8.3 g Na are also produced?</em>
- We need to get the no. of moles of 8.3 g Na using the relation:
n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.
- We can get the no. of moles of N₂ produced with 0.36 mol of Na:
<em><u>using cross multiplication:</u></em>
2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.
0.36 moles of Na produced with → ??? moles of N₂.
∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.
- We can get the mass of 0.54 mol of N₂:
mass = no. of moles x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.
- Now, we can get the mL of 15.12 g of N₂:
<em><u>using cross multiplication:</u></em>
1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.
??? L of N₂ contains → 15.12 g of N₂.
<em>∴ The volume of N₂ result </em>= (1.0 L)(15.12 g)/(0.92 g) = <em>16.434 L = 16434 mL.</em>
