Answer:
(1) mass of KCl =108.025g
(2) mass of MnCl2 =182.7 g
(3) mass of Cl2 =257.37 g
(4) mass of H20(water) =104.4g
Explanation:
we start with determining the limiting factor
molar mass of KMnO4 = 158g/mol
molar mass of HCl = 36.5g/mol
hence,
number of moles of KMnO4= mass /molar mass
number of moles of KMnO4= 229.19/158 = 1.45moles
number of moles of HCl = 526.64/36.5 = 14.43 moles
we chose the lowest number of moles from the reactants as the limiting factor
hence the limiting factor is KMnO4.
calculating the mass of the products:
for KCl
2moles of KMnO4 reacts to give 2moles of KCl
there for KCl contains 1.45moles
(1) mass of KCl = number of moles of KCl x molar mass of KCl = 1.45 x 74.5 = 108.025g
(2) from mole ratio 2 mole of KMnO4 gave 2 moles of MnCl2
therefore 1.45moles KMnO4 will give 1.45 MnCl2
mass of MnCl2 = 1.45moles x molar mass MnCl2 = 1.45 x 126 = 182.7 g
(3) from mole ratio
2moles KMnO4 gave 5moles Cl2
1.45 moles will give (5 x1.45)/2 moles of Cl2
mass of Cl2 = number of moles of Cl2 x molar mass Cl2 = 3.625 x 71 = 257.37 g
(4)from mole ratio
2moles KMnO4 gave 8moles water(H2O)
1.45 moles will give (1.45x 8)/2 moles of H20
mass of H2O produced = number of moles of H20 x the molar mass of H20 = 5.8 x 18 = 104.4g