Đáp án:
\(C\%_{H_2SO_4}=16,17\%; C\%_{CaSO_4}=4,76\%\)
Giải thích các bước giải:
\(n_{CaO}=\dfrac{2,8}{56}=0,05\ \text{mol}\\ n_{H_2SO_4}=\dfrac{140\times 20\%}{98}=\dfrac 27\ \text{mol}\\ CaO+H_2SO_4\to CaSO_4+H_2O\)
vì \(\dfrac{0,05}1<\dfrac27\to\)sau phản ứng CaO hết, \(H_2SO_4\) dư
\(n_{CaSO_4}=n_{CaO}=0,05\ \text{mol}\to m_{CaSO_4}=136\cdot 0,05=6,8\ \text{gam}\\ n_{H_2SO_4\ \text{p/u}}=n_{CaO}=0,05\ \text{mol}\to n_{H_2SO_4\ du}=\dfrac27-0,05=\dfrac{33}{140}\ \text{mol}\to m_{H_2SO_4\ dư}=98\cdot \dfrac{33}{140}=23,1\ \text{gam}\)
BTKL: \(\to m_{dd\ spu}=2,8+140=142,8\ \ \text{gam}\\\to C\%_{H_2SO_4}=\dfrac{23,1}{142,8}\cdot 100\%=16,17\%\\ C\%_{CaSO_4}=\dfrac{6,8}{142,8}\cdot 100\%=4,76\%\)
