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What is the limiting reactant when 45.9g of CuO are exposed to 2.57 g of H2 according to the equation CuO(s) + H2(g)−→Cu(s) + H...Question
What is the limiting reactant when 45.9g of CuO are exposed to 2.57 g of H2 according to the equation
CuO(s) + H2(g)−→Cu(s) + H2O(g) ?
1.CuO
2.Unable to determine.
3.H2
4.Neither is limiting.
CuO(s) + H2(g)−→Cu(s) + H2O(g) ?
1.CuO
2.Unable to determine.
3.H2
4.Neither is limiting.
Answers
MathMate
From the ratio of molar masses according to the equation, find the required mass of H2 for 45.9 g of CuO.
If the required mass of H2 is greater than 2.57g, then the limiting reagent is H2. If the required mass of H2 is less than 2.56g, then the limiting reagent is CuO.
MM for CuO=79.545
MM for H2 =2.016
mass of H2 required for 45.9 g of CuO
= (2.016/79.545)*45.9
= 1.16 g
Draw conclusion regarding which compound is the limiting reagent.
If the required mass of H2 is greater than 2.57g, then the limiting reagent is H2. If the required mass of H2 is less than 2.56g, then the limiting reagent is CuO.
MM for CuO=79.545
MM for H2 =2.016
mass of H2 required for 45.9 g of CuO
= (2.016/79.545)*45.9
= 1.16 g
Draw conclusion regarding which compound is the limiting reagent.
Farrell
Thanks friend!
MathMate
You're welcome! :)