when 56.6grams of Ca and 30.5grams Nitrogen gas undergo a reaction that has 9o% yield. what mass of calcium nitrate is formed?
First, we need to determine the balanced chemical equation for the reaction of calcium with nitrogen gas to form calcium nitrate.
Ca + N2 -> Ca3N2 (this is the reaction of calcium with nitrogen gas to form calcium nitride)
Next, we need to determine the balanced chemical equation for the reaction of calcium nitride with water to form calcium nitrate.
Ca3N2 + 6H2O -> 3Ca(NO3)2 + 2NH3 (this is the reaction of calcium nitride with water to form calcium nitrate and ammonia)
Now we can calculate the limiting reactant initially.
1. Convert grams to moles.
Moles of Ca = (56.6 g) / (40.08 g/mol) = 1.411 moles
Moles of N2 = (30.5 g) / (28.02 g/mol) = 1.088 moles
2. Determine the mole ratio of Ca3N2 that can be formed.
For every 3 moles of Ca, 1 mole of N2 can react.
(1.411 moles Ca) / 3 = 0.470 moles Ca
Since 1.088 moles of N2 > 0.470 moles of N2, the limiting reactant is Ca. So, 0.940 moles of Ca3N2 can be formed initially.
3. Calculate the theoretical yield of calcium nitrate.
1 mole of Ca3N2 produces 3 moles of Ca(NO3)2
0.940 moles Ca3N2 × 3 = 2.82 moles of Ca(NO3)2
Mass of Ca(NO3)2 = (2.82 moles) * (164.1 g/mol) = 462.95 g
4. Calculate the actual yield with 90% efficiency.
Actual yield = (90/100) × 462.95 g = 416.65 g
So, the mass of calcium nitrate formed is 416.65 grams with 90% yield.