Solution.
V(NaOH)=18.09mL=18.09⋅10−3L;
Concentration of NaOH =0.2235mol/L;
Amount of NaOH (mol) = Concentration of NaOH (mol/L) • Volume of NaOH consumed (L);
Amount of NaOH=0.2235mol/L⋅18.09⋅10−3L=4.043⋅10−3mol;
x mol 4.043⋅10−3mol
H2C2O4+2NaOH→Na2C2O4+2H2O;
1 2
Amount of H2C2O4=21⋅4.043⋅10−3mol=2.0215⋅10−3mol;
ν(H2C2O4)=M(H2C2O4)m(H2C2O4)⟹
m(H2C2O4)=ν(H2C2O4)M(H2C2O4);
M(H2C2O4)=90g/mol;
m(H2C2O4)=2.0215⋅10−3mol⋅90g/mol=0.182g;
Answer: m(H2C2O4)=0.182g.
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