Solution.
"V(NaOH)=18.09mL=18.09\\sdot10^{-3}L;"
Concentration of NaOH ="0.2235 mol\/L;"
Amount of NaOH (mol) = Concentration of NaOH (mol/L) • Volume of NaOH consumed (L);
Amount of NaOH="0.2235mol\/L\\sdot18.09\\sdot10^{-3}L=4.043\\sdot10^{-3}mol;"
x mol "4.043\\sdot10^{-3}mol"
"H_2C_2O_4+2NaOH\u2192Na_2C_2O_4+2H_2O;"
1 2
Amount of "H_2C_2O_4=\\dfrac{1\\sdot4.043\\sdot10^{-3}mol}{2}=2.0215\\sdot10^{-3}mol;"
"\\nu(H_2C_2O_4)=\\dfrac{m(H_2C_2O_4)}{M(H_2C_2O_4)}\\implies"
"m(H_2C_2O_4)=\\nu(H_2C_2O_4)M(H_2C_2O_4);"
"M(H_2C_2O_4)=90g\/mol;"
"m(H_2C_2O_4)=2.0215\\sdot10^{-3}mol\\sdot90g\/mol=0.182g;"
Answer: "m(H_2C_2O_4)=0.182g."
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