n(NaOH) = C * V = 1 mol;
m(NaOH) = n(NaOH) * M(NaOH) = 1 * 40 = 40 g;
2NaOH + CO 2 = Na 2 CO 3 + H 2 O; n(CO 2 )≥n(NaOH);
If n(CO 2 ) = 2 mol;
CaCO 3 = CaO + CO 2 ;
n(CaCO 3 ) = n(CO 2 ) = 2 mol;
m(CaCO 3 ) = n(CaCO 3 ) * M(CaCO 3 ) = 2*100 = 200 g;
m(limestone) = m(CaCO 3 )/ w(CaCO 3 ) = 200/0.6 = 333 g.
Answer: 333 g.
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