One mole of calcium carbide was treated with water, and $20.8*g$ of gas were collected. What is the percentage yield?

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Answer 1 · 2016-02-06T20:33:08

Approx. $80%$

$CaC-=C(s) + 2H_2O(l) rarr Ca(OH)_2(s) + HC-=CH(g)uarr$

$"Yield" = ("Moles of product")/("Moles of reactant") xx 100%$

$"Moles of product"$ $=$ $(20.8*cancelg)/(26.04*cancelg*mol^-1)$ $=$ $0.799$ $mol$

$"Moles of reactant"$ $=$ $1$ $mol$

$"Yield"$ $=$ $79.9%$

One mole of calcium carbide was treated with water, and 20.8*g20.8*g of gas were collected. What is the percentage yield?