One mole of calcium carbide was treated with water, and #20.8*g# of gas were collected. What is the percentage yield? Chemistry Stoichiometry Percent 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%# Answer link Related questions Question #e2e62 Question #fee2e Question #fee32 Question #19e89 Question #c5bdc Question #a1065 What would be the limiting reagent if 26.0 grams of C3H9N were reacted with 46.3 grams of... Is percent yield always less than 100? What would be the limiting reagent if 41.9 grams of C2H3OF were reacted with 61.0 grams of... What is the percent yield of the following reaction if 60 grams of CaCO3 is heated to give 15... See all questions in Percent Yield Impact of this question 1730 views around the world You can reuse this answer Creative Commons License