Answer:
We need 92.3 grams of sodium azide
Explanation:
Step 1: Data given
Mass of nitrogen gas = 59.6 grams
Molar mass of nitrogen gas = 28.0 g/mol
Molar mass of sodium azide = 65.0 g/mol
Step 2: The balanced equation
2NaN3 → 2Na + 3N2
Step 3: Calculate moles nitrogen gas
Moles N2 = mass N2 / molar mass N2
Moles N2 = 59.6 grams/ 28.0 g/mol
Moles N2 = 2.13 moles
Step 4: Calculate moles NaN3
for 2 moles NaN3 we'll have 2 moles Na and 3 moles N2
For 2.13 moles N2 we need 2/3* 2.13 = 1.42 moles NaN3
Step 5: Calculate mass NaN3
Mass NaN3 = Moles NaN3 * molar mass NaN3
Mass NaN3 = 1.42 moles * 65.0 g/mol
Mass NaN3 = 92.3 grams
We need 92.3 grams of sodium azide
There are several information's already given in the question. Based on those information's the answer can be easily determined.
M = <span>1.11 g CH4
C = </span>4.319 kJ g C⋅°
∆T = 35.65 - 24.85 degree centigrade
= 10.8 degree centigrade.
Then
∆H =−M ⋅ C ⋅∆T
= - 1.11 * 4.319 * 10.8
= - 51.776 kJ/<span>mol
I hope the procedure is clear enough for you to understand.</span>
