If 22.3mol of nitrogen is produced in the following reaction, how many moles of sodium is also produced? Note: The balancing is not correct. Use the coefficients provided even though they are not the right ones.
2Na3N → N2 + 6Na
well, using the equation,
1 mole of N2 and 6 moles of Na are produced.
That means six times as many moles of Na are produced as N2.
To determine the number of moles of sodium produced, we need to use the balanced equation for the reaction. However, it is mentioned that the balancing is not correct, so we will use the given coefficients.
The given balanced equation is:
2Na3N → N2 + 6Na
From the balanced equation, we can see that 1 mole of nitrogen gas (N2) is produced for every 2 moles of sodium azide (Na3N) consumed.
Given that 22.3 moles of nitrogen gas is produced, we can set up a proportion to find the number of moles of sodium.
Proportion: 2 mol Na3N / 1 mol N2 = x mol Na3N / 22.3 mol N2
Cross-multiplying and solving for x, we get:
x = (2 mol Na3N / 1 mol N2) * 22.3 mol N2
x = 44.6 mol Na3N
Therefore, 44.6 moles of sodium azide (Na3N) are produced in the given reaction.