Lithium metal reacts with nitrogen gas to produce lithium nitride .
What volume of nitrogen gas at 2.00 atm and 175 $\omicron $ C is required to produce 75.0 g of lithium nitride?
A) 30.9 L
B) 119 L
C) 79.2 L
D) 39.6 L
E) 15.5 L

Question

Quizplus · Accepted Answer

First, convert 175°F to °C: $175°F = \frac{5}{9}(175-32) = 79.44°C$. Then, convert °C to K: $79.44°C + 273.15 = 352.59K$. The balanced chemical equation for the reaction is $6Li + N_2 \rightarrow 2Li_3N$. The molar mass of $Li_3N$ is $3(6.94) + 14.01 = 34.83g/mol$. For 75.0 g of $Li_3N$, moles of $Li_3N = \frac{75.0g}{34.83g/mol} = 2.15mol$. From the balanced equation, 1 mole of $N_2$ produces 2 moles of $Li_3N$, so moles of $N_2 = \frac{2.15mol}{2} = 1.075mol$. Using the ideal gas law $PV = nRT$, where $P = 2.00 atm$, $n = 1.075 mol$, $R = 0.0821 \frac{L·atm}{mol·K}$, and $T = 352.59K$, solve for $V$: $V = \frac{nRT}{P} = \frac{(1.075mol)(0.0821\frac{L·atm}{mol·K})(352.59K)}{2.00atm} = 39.6 L$.

Lithium Metal Reacts with Nitrogen Gas to Produce Lithium Nitride ο\omicron ο

Lithium Metal Reacts with Nitrogen Gas to Produce Lithium Nitride $\omicron$