It\'s probably a calculation error but at this point I don\'t want to waste any
ID: 714414 • Letter: I
Question
It's probably a calculation error but at this point I don't want to waste any more attempts.
How many moles of N2 are produced by the decomposition of 1.10 mol of NaNs? n1.65 mol Correct Part B How many grams of NaNs are required to form 14.0 g of nitrogen gas? m- 32.5 Submit Previous Answers Request Answer Incorrect, Try Again; 5 attempts remaining Part C How many grams of NaNs are required to produce 11.0 ft of nitrogen gas if the gas has a density of 1.25 g/L? m- 389.35 Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remainingExplanation / Answer
B)
Molar mass of N2 = 28.02 g/mol
mass of N2 = 14 g
mol of N2 = (mass)/(molar mass)
= 14/28.02
= 0.4996 mol
Balanced chemical equation is:
2 NaN3 —> 2 Na + 3 N2
According to balanced equation
mol of NaN3 reqyured = (2/3)* moles of N2
= (2/3)*0.4996
= 0.3331 mol
Molar mass of NaN3,
MM = 1*MM(Na) + 3*MM(N)
= 1*22.99 + 3*14.01
= 65.02 g/mol
mass of NaN3 = number of mol * molar mass
= 0.3331*65.02
= 21.7 g
Answer: 21.7 g
C)
1 ft = 30.48 cm
volume = 11 ft^3
= 11 * (30.48 cm)^3
= 3.115*10^5 cm^3
= 3.115*10^5 mL
= 3.115*10^2 L
mass of N2 = density * volume
= 1.25 g/L * 3.115*10^2 L
= 389.4 g
Molar mass of N2 = 28.02 g/mol
mass of N2 = 3.894*10^2 g
mol of N2 = (mass)/(molar mass)
= 3.894*10^2/28.02
= 13.9 mol
Balanced chemical equation is:
2 NaN3 —> 2 Na + 3 N2
According to balanced equation
mol of NaN3 required = (2/3)* moles of N2
= (2/3)*13.9
= 9.265 mol
Molar mass of NaN3,
MM = 1*MM(Na) + 3*MM(N)
= 1*22.99 + 3*14.01
= 65.02 g/mol
mass of NaN3 = number of mol * molar mass
= 9.265*65.02
= 602 g
Answer: 602 g
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