Limiting Reactant Problems

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Limiting Reactant Problems

Postby Maldonado3K » Sun Apr 08, 2018 11:57 pm

In Lavelle's last Friday example, can someone explain in simpler steps how to solve the last part (finding the maximum yield)?

Isabel Jabara 1C
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Joined: Fri Apr 06, 2018 11:03 am

Re: Limiting Reactant Problems

Postby Isabel Jabara 1C » Mon Apr 09, 2018 6:11 pm

After finding the limiting reactant, you must look at the ratio of the limiting reactant to the product. Since the ratio is 1 mol CaC2 to 1 mol CaC2, the maximum yield of 1.56 mol CaC2 is 1.56 mol C2H2. Then you multiply this by the molar mass of C2H2 which is 26.038 g/mol.

1.56 mol x 26.038 g/mol = 40.6 g

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