\(\begin{array}{l}
1)\\
mKCl\,trong\,{\rm{dd}}\,10\%  = 200 \times 10\%  = 20g\\
m{H_2}O = 200 - 20 = 180g\\
mKCl\,trong\,{\rm{dd}}bh = \dfrac{{180 \times 31,2}}{{100}} = 56,16g\\
m = 56,16 - 20 = 36,16g\\
2)\\
m{K_2}C{r_2}{O_7} = \dfrac{{200 \times 6,54}}{{100}} = 13,08g\\
m{H_2}O = 200 - 13,08 = 186,92g\\
S = \dfrac{{13,08}}{{186,92}} \times 100 = 7g\\
3)
\end{array}\)

\(\begin{array}{l}
mKN{O_3}\,trong\,{\rm{dd}}\,10\%  = 10\%  \times a = 0,1a\,(g)\\
m{H_2}O = a - 0,1a = 0,9a(g)\\
mKN{O_3}\,trong\,{\rm{dd}}\,bh = \dfrac{{0,9a \times 21,9}}{{100}} = 0,1971a(g)\\
a + m = 200(1)\\
0,1a + m = 0,1971a(2)\\
(1) - (2) \Rightarrow 0,9a = 200 - 0,1971a \Rightarrow a = 182,3g\\
m = 200 - 182,3 = 17,7g
\end{array}\)

Đáp án:

Giải thích các bước giải:

1)mKCltrongdd10%=200×10%=20gmH2O=200−20=180gmKCltrongddbh=180×31,2100=56,16gm=56,16−20=36,16g2)mK2Cr2O7=200×6,54100=13,08gmH2O=200−13,08=186,92gS=13,08186,92×100=7g3)1)mKCltrongdd10%=200×10%=20gmH2O=200−20=180gmKCltrongddbh=180×31,2100=56,16gm=56,16−20=36,16g2)mK2Cr2O7=200×6,54100=13,08gmH2O=200−13,08=186,92gS=13,08186,92×100=7g3)

mKNO3trongdd10%=10%×a=0,1a(g)mH2O=a−0,1a=0,9a(g)mKNO3trongddbh=0,9a×21,9100=0,1971a(g)a+m=200(1)0,1a+m=0,1971a(2)(1)−(2)⇒0,9a=200−0,1971a⇒a=182,3gm=200−182,3=17,7g