Đáp án:
\( {m_{BaS{O_4}}} = 23,3{\text{ gam}}\)
\( C{\% _{NaCl}} = 5,16\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(N{a_2}S{O_4} + BaC{l_2}\xrightarrow{{}}BaS{O_4} + 2NaCl\)
Ta có:
\({n_{N{a_2}S{O_4}}} = \frac{{14,2}}{{23.2 + 96}} = 0,1{\text{ mol = }}{{\text{n}}_{BaS{O_4}}}\)
\( \to {m_{BaS{O_4}}} = 0,1.(137 + 96) = 23,3{\text{ gam}}\)
\({m_{dd}} = {m_{dd{\text{ N}}{{\text{a}}_2}S{O_4}}} + {m_{dd\;{\text{BaC}}{{\text{l}}_2}}} - {m_{BaS{O_4}}}\)
\( = 100 + 150 - 23,3 = 226,7{\text{ gam}}\)
\({n_{NaCl}} = 2{n_{N{a_2}S{O_4}}} = 0,2{\text{ mol}}\)
\( \to {m_{NaCl}} = 0,2.58,5 = 11,7{\text{ gam}}\)
\( \to C{\% _{NaCl}} = \frac{{11,7}}{{226,7}} = 5,16\% \)
