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rizwan440 rizwan440
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12 years ago
For each of the following unbalanced reactions, suppose 5 g of each reactant is taken. Determine which reactant is limiting, and also determine what mass of the excess reagent will remain after the limiting reactant is consumed.

CaC2(s) + H2O(l)  Ca(OH)2(s) + C2H2(g)


-thankS!

Can you show me the work too.
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#1
12 years ago
Let UNITs guide you; always USE THEM in your calculation to prevent errors

Firstly, the eqn: CaC2(s) + 2H2O(l) -------> Ca(OH)2(s) + C2H2(g)

moles CaC2 = 5g/ MW CaC2 g/mole = ??
moles H2O = 5 g/ MW H2O g/mole = ??

IF, moles CaC2 > 2 * moles H2O, then H2O is limiting; otherwise CaC2 is

excess moles of H2O = moles H2O init - 2*moles CaC2 = ??

g H2O excess = 18 g/mole * excess moles of H2O = ??

Plug and SOLVE

Basic mathematics is a prerequisite to chemistry ? I just try to help you with the methodology of solving the problem.
wrote...
#2
12 years ago
CaC2 has a higher MW than does water.
Therefore there will be fewer moles of it in 5 g.
Therefore it will be the limiting reagent.
Water will be present in excess

So to get an exact answer, find MW CaC2. Find how many moles in 5 g. Do the same for water. Subtract. Difference is how many moles of water remain. Convert back into grams.
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