Need help with this question, Could you also explain how it\'s done as well? Tha

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Need help with this question, Could you also explain how it's done as well? Thanks!
11. Cyanogen gas, C2N2, has been found in outer space gases. It can react with fluorine to form carbon tetrafluoride and nitrogen trifluoride following the reaction below: C2N2 (g) + 7 F2 (g) 2 CF4 (g) + 2 NF3 (g) a. How many moles of fluorine react with 1.37 moles of cyanogen? b. How many moles of CF4 are obtained from 13.75 moles of fluorine? c. How many moles of cyanogen are required to produce 0.8974 moles of NF3? d. How many moles of fluorine will yield 4.981 moles of nitrogen trifluoride? Need help with this question, Could you also explain how it's done as well? Thanks!
11. Cyanogen gas, C2N2, has been found in outer space gases. It can react with fluorine to form carbon tetrafluoride and nitrogen trifluoride following the reaction below: C2N2 (g) + 7 F2 (g) 2 CF4 (g) + 2 NF3 (g) a. How many moles of fluorine react with 1.37 moles of cyanogen? b. How many moles of CF4 are obtained from 13.75 moles of fluorine? c. How many moles of cyanogen are required to produce 0.8974 moles of NF3? d. How many moles of fluorine will yield 4.981 moles of nitrogen trifluoride? Need help with this question, Could you also explain how it's done as well? Thanks!
11. Cyanogen gas, C2N2, has been found in outer space gases. It can react with fluorine to form carbon tetrafluoride and nitrogen trifluoride following the reaction below: C2N2 (g) + 7 F2 (g) 2 CF4 (g) + 2 NF3 (g) a. How many moles of fluorine react with 1.37 moles of cyanogen? b. How many moles of CF4 are obtained from 13.75 moles of fluorine? c. How many moles of cyanogen are required to produce 0.8974 moles of NF3? d. How many moles of fluorine will yield 4.981 moles of nitrogen trifluoride?

Explanation / Answer

a.

from the reaction

we need 1 mol of C2N2 per 7 mol of F2

therefore, ratio is 1:7, or 7x

then

1.37 mol need = 1.37*7 = 9.59 mol of F2

b.

7 mol of F2 forms 2 mol of CF4, then ratio is 2/7

13.75 mol of F2 --> 13.75*2/7 = 3.928 mol of CF4

c.

1 mol of C2N2 --> 2 mol of NF3

ratio is 1/2

then

0.8974 --> 1/2*0.8974 = 0.4487 mol of NF3

d.

ratio from

2 NF3 = 7 F2

7/2

then

4.981*7/2 = 17.4335 mol o fF2 needed

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