0.00796mol of CaC2 react with 8.324mol of H2O to produce C2H2 and Ca(OH)2.
Equation:
CaC2 (s) + 2 H2O (l) → C2H2 (g) + Ca(OH)2 (aq)
Calculate the actual yield of C2H2(g) and the actual yield of Ca(OH)2
Equation:
CaC2 (s) + 2 H2O (l) → C2H2 (g) + Ca(OH)2 (aq)
Calculate the actual yield of C2H2(g) and the actual yield of Ca(OH)2
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This is a limiting reagent problem. Which reactant will be used up before the other.
The mol ration between CaC2 and H2O is 1 to 2 according to the equation. For every mol of H2O that is used up you only need 1/2 that many moles of CaC2
You have 8.324 mol of H2O so you will need at least 1/2 x 8.324 or 4.162 mol of CaC2 to use up all the water.
Well, you clearly don't have that much, so the reaction will stop completely when the 0.00796 mol of CaC2 is all used up.
The mol ration between CaC2, C2H2 and Ca(OH)2 are all 1 : 1 : 1
So if you use up 0.00796 mol of CaC2 you will produce 0.00796 mol of C2H2 and Ca(OH)2 as well.
The mol ration between CaC2 and H2O is 1 to 2 according to the equation. For every mol of H2O that is used up you only need 1/2 that many moles of CaC2
You have 8.324 mol of H2O so you will need at least 1/2 x 8.324 or 4.162 mol of CaC2 to use up all the water.
Well, you clearly don't have that much, so the reaction will stop completely when the 0.00796 mol of CaC2 is all used up.
The mol ration between CaC2, C2H2 and Ca(OH)2 are all 1 : 1 : 1
So if you use up 0.00796 mol of CaC2 you will produce 0.00796 mol of C2H2 and Ca(OH)2 as well.