Question

How many grams of lithium are required to completely react with 61.3 mL of N2 gas at STP?Express your answer in grams to three significant figures.

Answer 1

0.114 g of lithium are necessary to react with 61.3 mL of N2.

First, we need the balanced reaction to work on this problem. When lithium metal (Li) reacts with nitrogen gas (N2), it forms Li3N:

6Li + N2 --> 2Li3N

Under Standard Temperature and Pressure (STP), 1 mol of any compound corresponds to 22.4L of that compound:

1 mol N2 ------ 22.4 L

x mol N2 -------- 61.3 x 10^-3 L

Then, solving for x, we have that there are 0.00274 mol of N2 in 61.3 mL of N2.

Now, we can calcute the amount of Li necessary to react with 0.00274 mol of N2 and then convert this value into grams using the Li atomic mass (6.941 u):

6 mol Li ------ 1 mol N2

y --------------- 0.00274 mol N2

Solving for y, we have 0.0164 mol N2

Since there are 6.941 g in 1 mol of this compound:

1 mol Li ---------- 6.941 g Li

0.01644 mol Li ---------- z

Solving for z, we have 0.114 g of lithium.