Question

Cyanogen, ${\left(\mathrm{CN}\right)}_{\mathbf{2}}$ has been observed in the atmosphere of Titan, Saturn’s largest moon, and in the gases of interstellar nebulas. On Earth, it is used as a welding gas and a fumigant. In its reaction with fluorine gas, carbon tetrafluoride and nitrogen trifluoride gases are produced. What mass of carbon tetrafluoride forms when 60.0 g of each reactant is used?

Short answer

Mass formed of ${\mathbf{CF}}_{\mathbf{4}}$during the reaction is 39.6 g.

Step-by-step solution

Writing balanced equation

First of all, let us check the balanced equation to find the amount of${\mathbf{CF}}_{\mathbf{4}}$

${\left(\mathrm{CN}\right)}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{7}{\mathbf{F}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{2}{\mathbf{CF}}_{\mathbf{4}}\left(g\right)\mathbf{+}\mathbf{2}{\mathbf{NF}}_{\mathbf{3}}\left(g\right)$

From the equation we can say that 1 mole of${\left(\mathrm{CN}\right)}_{\mathbf{2}}$and 7 moles of fluorine gasreact to givetwomoles of ${\mathbf{CF}}_{\mathbf{4}}$

Calculating Moles of reactants

Moles can be calculated as

$\mathbf{Mole}\mathbf{}\mathbf{of}\mathbf{}{\left(\mathrm{CN}\right)}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{Mole}\mathbf{}\mathbf{of}\mathbf{}{\left(\mathrm{CN}\right)}_{\mathbf{2}}}{\mathbf{Molar}\mathbf{}\mathbf{Mass}}\mathbf{=}\frac{\mathbf{60}}{\mathbf{92}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{65}$

$\mathbf{Mole}\mathbf{}{\mathbf{ofF}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{Mole}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{F}}_{\mathbf{2}}}{\mathbf{Molar}\mathbf{}\mathbf{Mass}}\mathbf{=}\frac{\mathbf{60}}{\mathbf{38}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{58}$

From the balanced equation we can say that${\mathbf{F}}_{\mathbf{2}}$is the limiting reagent hence controls the formation of products and,

7 moles of${\mathbf{F}}_{\mathbf{2}}$gives moles of${\mathbf{CF}}_{\mathbf{4}}$= 2 moles

So, 1.56 moles of ${\mathbf{F}}_{\mathbf{2}}$ will give = $\frac{\mathbf{2}}{\mathbf{7}}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{58}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{45}\mathbf{Mole}$

Determine mass of  CF4 formed 

The mass of${\mathbf{CF}}_{\mathbf{4}}$formed is calculated as:

$\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{CF}}_{\mathbf{4}}\mathbf{}\mathbf{=}\mathbf{Mole}\mathbf{}\mathbf{\times }\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{45}\mathbf{\times }\mathbf{88}\mathbf{=}\mathbf{39}\mathbf{.}\mathbf{6}\mathbf{g}$

So, mass formed of ${\mathbf{CF}}_{\mathbf{4}}$ during the reaction is 39.6 g.