Đáp án:
\(\begin{array}{l}
a)\\
{m_{N{a_2}S{O_4}}} = 28,4g\\
{m_{KN{O_3}}} = 30,3g\\
b)\\
{m_{BaC{l_2}pu}} = 41,6g\\
c)\\
{m_{BaC{l_2}}} = 49,92g
\end{array}\)
Giải thích các bước giải:
a)
\(\begin{array}{l}
BaC{l_2} + N{a_2}S{O_4} \to BaS{O_4} + 2NaCl\\
{n_{BaS{O_4}}} = \dfrac{{46,6}}{{233}} = 0,2mol\\
{n_{N{a_2}S{O_4}}} = {n_{BaS{O_4}}} = 0,2mol\\
{m_{N{a_2}S{O_4}}} = 0,2 \times 142 = 28,4g\\
{m_{KN{O_3}}} = 58,7 - 28,4 = 30,3g\\
b)\\
{n_{BaC{l_2}pu}} = {n_{BaS{O_4}}} = 0,2mol\\
{m_{BaC{l_2}pu}} = 0,2 \times 208 = 41,6g\\
c)\\
{m_{BaC{l_2}d}} = \dfrac{{41,6 \times 20}}{{100}} = 8,32g\\
{m_{BaC{l_2}}} = 41,6 + 8,32 = 49,92g
\end{array}\)