Đáp án:
a) m dung dịch HNO3=5,25 gam
b) C% NaNO3=5,86 %
Giải thích các bước giải:
Ta có: \({m_{NaOH}} = 200.10\% = 20{\text{ gam}} \to {{\text{n}}_{NaOH}} = \frac{{20}}{{40}} = 0,5{\text{ mol}}\)
\(NaOH + HN{O_3}\xrightarrow{{}}NaN{O_3} + {H_2}O\)
\(\to {n_{HN{O_3}}} = {n_{NaN{O_3}}} = {n_{NaOH}} = 0,5{\text{ mol}} \to {{\text{m}}_{HN{O_3}}} = 0,5.63 = 31,5{\text{ gam}} \to {{\text{m}}_{dd{\text{ }}HN{O_3}}} = \frac{{31,5}}{{6\% }} = 525{\text{ gam}}\)
BTKL: \({m_{dd\;{\text{muối}}}} = 200 + 525 = 725{\text{ gam}}\)
\({m_{NaN{O_3}}} = 0,5.(23 + 62) = 42,5{\text{ gam}} \to {\text{C}}{{\text{\% }}_{NaN{O_3}}} = \frac{{42,5}}{{725}} = 5,86\% \)
