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

1 Answer
anor277
Feb 6, 2016

Approx. #80%#

Explanation:

#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%#

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