CaF2 + H2SO4 → 2HF + CaSO4
M(CaF2) = 78.07 g/mol
M(H2SO4) = 98.08 g/mol
M(HF) = 20.01 g/mol
n = m/M
n(CaF2) "= \\frac{10.0}{78.07} = 0.128 \\;mol"
n(H2SO4)"= \\frac{15.5}{98.08} = 0.158 \\;mol"
Proportion according to the reaction:
1 mol CaF2 : 1 mol H2SO4
Actual amount of reactants:
0.128 mol CaF2 : 0.158 mol H2SO4
So, H2SO4 is in excess and CaF2 is the limiting reactant.
n(HF) = 2n(CaF2) "= 2 \\times 0.128 = 0.256 \\;mol"
m(CaF2) "= 0.256 \\times 20.01 = 5.12 \\;g"
n(H2SO4) = 0.128 mol (needs for reaction)
∆n(H2SO4) = 0.158 – 0.128 = 0.03 mol
Mass of excess reactant:
m(H2SO4) "= 0.03 \\times 98.08 = 2.94 \\;g"
Comments
Leave a comment