\(n_{K2CO3}=\dfrac{200.13,8\%}{100\%.138}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{100.18,25\%}{100\%.36,5}=0,5\left(mol\right)\)
a) Pt : \(K_2CO_3+HCl\rightarrow KCl+CO_2+H_2O|\)
0,2 0,5 0,2 0,2
Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{0,5}{1}\Rightarrow K_2CO3hết,HCldư\)
\(V_{CO2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
b) \(m_{KCl}=0,2.74,5=14,9\left(g\right)\)
\(m_{NaOH\left(dư\right)}=\left(0,5.0,4\right).40=4\left(g\right)\)
\(m_{ddspu}=200+100-\left(0,2.44\right)=291,2\left(g\right)\)
\(C\%_{K2CO3}=\dfrac{14,9.100}{219,2}=5,12\%\)
\(C\%_{ddNaOH\left(dư\right)}=\dfrac{4.100}{291,2}=1,37\%\)
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