63.1 g of NaN3 is formed when 50.0 g of sodium is reacted with 40.5 g of nitrogen according to the balanced chemical reaction
The balanced reaction equation is;
2 Na + 3 N2--> 2 NaN3
Number of moles of Na = 50.0 g/23 g/mol = 2.17 moles of Na
Number of moles of Nitrogen = 40.5 g/28 g/mol = 1.45 moles of N2
We have to obtain the limiting reactant, this is the reactant that yields the least number of moles of product.
For Na
2 moles of Na yields 2 moles of NaN3
2.17 moles of Na yields 2.17 moles of NaN3 (reaction is 1:1).
For N2
3 moles of N2 yields 2 moles of NaN3
1.45 moles of N2 yields 1.45 * 2/3 = 0.97 moles of NaN3
So, N2 is the limiting reactant. Mass of product formed depends on the limiting reactant.
Mass of NaN3 = 0.97 moles of NaN3 * 65 g/mol = 63.1 g of NaN3
Therefore, 63.1 g of NaN3 is formed when 50.0 g of sodium is reacted with 40.5 g of nitrogen according to the balanced chemical reaction
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