Nitrogen and fluorine react to form nitrogen fluoride according to the chemical

Question

Nitrogen and fluorine react to form nitrogen fluoride according to the chemical equation: N2 (8)+3F2(8)- 2NF3 (8) If a sample containing 36.0 g of N2 is reacted with 36.0 g of F2, how many grams of NFg will be produced? HOW DO WE GET THERE? Now we need to find the amount of NF3 that can be formed by the complete reactions of each of the reactant If all of the N2 was used up in the reaction, how many moles of NF3 would be produced? mol NF Next) (3 of 6) Check Submit Answer Try Another Version 10 item attempts remaining

Explanation / Answer

Molar mass of N2 = 28.02 g/mol


mass(N2)= 36.0 g

use:
number of mol of N2,
n = mass of N2/molar mass of N2
=(36.0 g)/(28.02 g/mol)
= 1.285 mol

Molar mass of F2 = 38 g/mol


mass(F2)= 36.0 g

use:
number of mol of F2,
n = mass of F2/molar mass of F2
=(36.0 g)/(38 g/mol)
= 0.9474 mol
Balanced chemical equation is:
N2 + 3 F2 ---> 2 NF3 +


1 mol of N2 reacts with 3 mol of F2
for 1.285 mol of N2, 3.854 mol of F2 is required
But we have 0.9474 mol of F2

so, F2 is limiting reagent
we will use F2 in further calculation


Molar mass of NF3,
MM = 1*MM(N) + 3*MM(F)
= 1*14.01 + 3*19.0
= 71.01 g/mol

According to balanced equation
mol of NF3 formed = (2/3)* moles of F2
= (2/3)*0.9474
= 0.6316 mol


use:
mass of NF3 = number of mol * molar mass
= 0.6316*71.01
= 44.85 g
Answer: 44.8 g

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