How can I identify the limiting reactant when 43.25 g of cac2 reacts with 33.71 g of water to produce ca (oh) to and c2h2?

1 Answer

Answer:

#"CaC"""_2# is the limiting reagent.

Explanation:

The limiting reagent is #CaC_2#.

First, start with the balanced chemical equation

#CaC_2 + 2H_2O -> C_2H_2 + Ca(OH)_2#

Notice that we have a #1:2# mole ratio between #CaC_2# and #H_2O#; that is, every mole of the former used requires 2 moles of the latter.

Since the quantities of both reactants are given, and knowing that their molar masses are #62.0 g/(mol)# (for #CaC_2#) and #18.0 g/(mol)# (for #H_2O#), we can determine the number of moles from

#n_(H_2O) = m_(H_2O)/(molarmass_(H_2O)) = (33.71 g)/(18 g/(mol)) = 1.87# moles

#n_(CaC_2) = m_(CaC_2)/(molarmass_(CaC_2)) = (43.25 g)/(64.0 g/(mol)) = 0.68# moles

Notice that the number of moles of #CaC_2# determined would require #2 * 0.68 = 1.36# moles of #H_2O#, less than what we have in the reaction. Therefore, #H_2O# is in excess, which means that #CaC_2# is the limiting reagent.