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?

Question

Owen Ambrose · Accepted Answer

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

The limiting reagent is #CaC_2#. First, utilize the chemical equation that is balanced. #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.

Axel Alvarado · Answer

undefined To identify the limiting reactant, first, calculate the moles of each reactant using their respective molar masses. Then, use the stoichiometry of the balanced chemical equation to determine the moles of products that each reactant can produce. The reactant that produces the lesser amount of product is the limiting reactant.