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
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.
To determine the limiting reactant, we need to compare the number of moles of each reactant.
Step 1: Convert the mass of each reactant to moles using their respective molar masses.
Molar mass of NaSO4 = 22.99 g/mol (Na) + 32.07 g/mol (S) + (4 * 16.00 g/mol) (O)
= 98.09 g/mol
Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + (3 * 16.00 g/mol) (O)
= 100.09 g/mol
Moles of NaSO4 = mass / molar mass
= 52.54 g / 98.09 g/mol
≈ 0.535 mol
Moles of CaCO3 = mass / molar mass
= 45 g / 100.09 g/mol
≈ 0.449 mol
Step 2: Divide the number of moles of each reactant by their respective stoichiometric coefficient in the balanced chemical equation.
The balanced chemical equation for the reaction is:
Na2SO4 + CaCO3 -> Na2CO3 + CaSO4
From the balanced equation, we can see that the stoichiometric coefficients are 1:1 for Na2SO4 and CaCO3.
Moles of NaSO4 / stoichiometric coefficient = 0.535 mol / 1 = 0.535 mol
Moles of CaCO3 / stoichiometric coefficient = 0.449 mol / 1 = 0.449 mol
Step 3: Compare the calculated values from step 2.
Based on the comparison, we can see that the limiting reactant is CaCO3 because it produces fewer moles of product compared to Na2SO4.