16 gram of CaC2 and 18 gram of H2O were mixed in the given reaction,
CaC2 + 2H2O it gives C2H2 + Ca(OH)2. what is the limiting reactant of the give equation?
To determine the limiting reactant, we need to calculate the moles of each reactant and compare them with their stoichiometric coefficients in the balanced equation.
Moles of CaC2 = 16 g / 64.1 g/mol = 0.25 mol
Moles of H2O = 18 g / 18.0 g/mol = 1.00 mol
From the balanced equation, we can see that 1 mole of CaC2 reacts with 2 moles of H2O. Therefore, the amount of H2O required for 0.25 moles of CaC2 to react completely is:
2 mol H2O / 1 mol CaC2 x 0.25 mol CaC2 = 0.50 mol H2O
Since we have 1.00 mol of H2O, this means that H2O is in excess and CaC2 is the limiting reactant.
Therefore, the limiting reactant is CaC2.