Question

Question: For the reaction\(_6^{14}C \to _7^{14}N + ? ,if 100.0\;g\)of carbon reacts, what volume of nitrogen gas\(\left( {{N_2}} \right)\)is produced at\(273\;K and 1\;atm\)?

Short answer

\(V\left( {{N_2}} \right) = 320.00\;mL\)

Step-by-step solution

The volume of nitrogen gas \(\left( {{N_2}} \right)\)

We have:\(_6^{14}C \to _7^{14}N + _y^x?\)

The mass on the left side is 14, which means we also need to have the sum of 14 on the right side:

\(\begin{array}{c}x + 14 = 14\\x = 0\end{array}\)

The atomic number on the left is 6, which means we also need to have the sum of 14 on the right side:

\(\begin{array}{c}y + 7 = 6\\y = - 1\end{array}\)

The particle which has an atomic number of\( - 1\)and mass number of 0 is an electron, which means the complete reaction is:

\(_6^{14}C \to _7^{14}N + _{ - 1}^0e\)

This reaction implies that 1 mole of carbon-14 reacts to form 1 mole of nitrogen-14.

The volume of nitrogen gas \(\left( {{N_2}} \right)\) produced at \(273\;K and 1\;atm\)

We have simple formula from which we can get the volume from mass, temperature, and pressure :

\(PV = nRT\)

Where\(R\)is a universal gas constant\(\left( {R = 8.314\frac{J}{{Kmol}}} \right).\)

The pressure of\(1\;atm\)is equal to\(101325\;Pa.\)

We are only missing is the number of moles of nitrogen gas\(\left( {{N_2}} \right),\)

Which is equal to the number of moles of carbon.

\(\begin{array}{c}\frac{{n(C)}}{{n(N)}} = \frac{1}{1}\\n(N) = n(C)\\\frac{m}{M} = \frac{{100\;g}}{{14gmo{l^{ - 1}}}}\\ = 7.143\;mol\end{array}\)

We have not 1 nitrogen atom in nitrogen gas, but 2

\(\begin{array}{c}n\left( {{N_2}} \right) = 2 \times 7.143\;mol\\ = 14.286\;mol\end{array}\)

Now we can get the volume:

\(\begin{array}{c}PV = nRT\\V\left( {{N_2}} \right) = \frac{{nRT}}{P}\\V\left( {{N_2}} \right) = \frac{{14.286\;mol \times 8.314\frac{J}{{Kmol}} \times 273\;K}}{{101325\;Pa}}\\V\left( {{N_2}} \right) = 0.32\;L\\ = 320.00\;mL\end{array}\)