How many grams of lithium metal are required to react with a #28*g# mass of dinitrogen gas?

Among the few metals that can directly react with nitrogen gas at room temperature is lithium. One VERY costly method of producing ammonia gas is to use lithium nitride:

Therefore, 42 g of lithium metal and 28 g of dinitrogen gas are needed for the reaction, given the stoichimetry.

To determine the grams of lithium metal required to react with 28 g of dinitrogen gas ((N_2)), we use stoichiometry.

The balanced chemical equation for the reaction is:

[ 6Li + N_2 \rightarrow 2Li_3N ]

The molar mass of dinitrogen gas ((N_2)) is 28 g/mol.

Using stoichiometry, we find the molar mass of lithium ((Li)) is approximately 6.94 g/mol.

To find the grams of lithium required, we need to set up a ratio based on the stoichiometry of the reaction:

[ \frac{6 \text{ mol Li}}{1 \text{ mol } N_2} = \frac{x \text{ g Li}}{28 \text{ g } N_2} ]

Solving for (x), we get:

[ x = \frac{6 \times 28}{1} = 168 \text{ g} ]

Therefore, 168 grams of lithium metal are required to react with 28 g of dinitrogen gas.