In closed container 52,54g of NaSO4 is allowed to react with 45g of CaCO3 in the presence of rough C . Determine the limiting the reactant

Answer 1

To determine the limiting reactant, we need to compare the moles of NaSO4 and CaCO3.

First, we need to calculate the molar mass of NaSO4 and CaCO3:
NaSO4: Na (22.99 g/mol) + S (32.07 g/mol) + 4O (16.00 g/mol) = 119.04 g/mol
CaCO3: Ca (40.08 g/mol) + C (12.01 g/mol) + 3O (16.00 g/mol) = 100.09 g/mol

Now we can calculate the moles of NaSO4 and CaCO3:
moles of NaSO4 = 52.54 g / 119.04 g/mol = 0.4411 mol
moles of CaCO3 = 45 g / 100.09 g/mol = 0.4497 mol

The limiting reactant is the one with the least number of moles. In this case, NaSO4 has the least number of moles (0.4411 mol) compared to CaCO3 (0.4497 mol). Therefore, NaSO4 is the limiting reactant.