Q87P - Chemistry: Molecular Nature Of Matter And Change

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?

Step-by-Step Solution

Answer

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

Step 1: 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  ${\mathbf{CF}}_{\mathbf{4}}$

Step 2: 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 =   $\frac{\mathbf{2}}{\mathbf{7}}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{58}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{45}\mathbf{Mole}$

Step 3: Determine mass of CF 4 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.