Respuesta :
% yield of SnS2 = 0.337 X 100 / 0.567 = 59.44 %
% yield of SnS2 = 59.44 %
Balanced chemical reaction is
SnBr4(aq) + 2Na2S(aq) \rightarrow 4NaBr(aq) + SnS2(s)
First calculate limiting reactant
molarity of SnBr4 = 0.418 M
volume of SnBr4 = 0.0451 L (1 ml = 0.001 L then 45.1 ml =45.1 X 0.001 = 0.0451 L)
molarity of Na2S = 0.133 M
volume of Na2S = 0.0466 M ( 1 ml = 0.001 L then 46.6 ml =46.6 X 0.001 = 0.0466 L)
no.of moles = molarity X volume of solution in liter
moles of SnBr4 = 0.418 X 0.0451 = 0.0189 mol
moles of Na2S = 0.133 X 0.0466 = 0.0062 mol
According to balanced chemical reaction 1 mole of SnBr4 react with 2 mole of Na2S molar ratio between SnBr4 to Na2S is 1:2 therefore to react with 0.0189 mole of SnBr4 required Na2S = 0.0189 X 2 / 1 = 0.0378 moles Na2S but Na2S given only 0.0062 mole therefore Na2S is limiting reactant
Limiting reactant = Na2S
According to balanced chemical reaction 2 mole of Na2S produce 1 mole of SnS2 molar ratio between Na2S to SnS2 is 2:1 therefore 0.0062 mole of Na2S produce SnS2 = 0.0062 X 1 / 2 = 0.0031 moles
Mole of SnS2 produced = 0.0031 mole
Molar mass of SnS2 = 182.81 g/mol
Gm of compound = no. of moles X molar mass
Gm of SnS2 formed = 0.0031 mol X 182.81 g/mol = 0.567 g
Gm of SnS2 formed = 0.567 g
Theoretical yield of SnS2 = 0.567 g
Actual yield of SnS2 = 0.337 g
Theoretical yield of SnS2 = 0.567 g
% yield = Actual yield X 100 / theoretical yield
Substitute the value
% yield of SnS2 = 0.337 X 100 / 0.567 = 59.44 %
% yield of SnS2 = 59.44 %
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