Đáp án:
a, `V_{Cl_2}=3,36\ (l).`
b, `m_{dd\ H_2SO_4}=210\ (g).`
c,
`C%_{KOH\ (dư)}=3,45%`
`C%_{KCl}=2,3%`
`C%_{KClO_3}=0,76%`
Giải thích các bước giải:
a,
`n_{NaCl}=\frac{300\times 5,85\%}{58,5}=0,3\ (mol).`
Phương trình hóa học:
`2NaCl + 2H_2O \overset{đpdd, cmn}\to 2NaOH + Cl_2\uparrow + H_2\uparrow`
`-` Theo pt: `n_{Cl_2}=\frac{1}{2}n_{NaCl}=0,15\ (mol).`
`\to V_{Cl_2}=0,15\times 22,4=3,36\ (l).`
`-` Theo pt: `n_{NaOH}=n_{NaCl}=0,3\ (mol).`
b,
Phương trình hóa học:
`2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O`
`-` Theo pt: `n_{H_2SO_4}=\frac{1}{2}n_{NaOH}=0,15\ (mol).`
`\to V_{dd\ H_2SO_4}=\dfrac{0,15}{1}=0,15\ (l)=150\ (ml).`
`\to m_{dd\ H_2SO_4}=150\times 1,4=210\ (g).`
c,
Phương trình hóa học:
`3Cl_2 + 6KOH \to 5KCl + KClO_3 + 3H_2O`
`\sum n_{KOH}=\frac{800\times 5,6\%}{56}=0,8\ (mol).`
Tỉ lệ: `n_{Cl_2}:n_{KOH}=\frac{0,15}{3}<\frac{0,8}{6}.`
`\to KOH` dư.
`\to n_{KOH\ (dư)}=0,8-(0,15\times 2)=0,5\ (mol).`
`-` Theo pt:
`n_{KCl}=\frac{5}{3}n_{Cl_2}=0,25\ (mol).`
`n_{KClO_3}=\frac{1}{3}n_{Cl_2}=0,05\ (mol).`
`m_{dd\ spư}=m_{Cl_2}+m_{dd KOH}=0,15\times 71+800=810,65\ (g).`
`\to C%_{KOH\ (dư)}=\frac{0,5\times 56}{810,65}\times 100%=3,45%`
`\to C%_{KCl}=\frac{0,25\times 74,5}{810,65}\times 100%=2,3%`
`\to C%_{KClO_3}=\frac{0,05\times 122,5}{810,65}\times 100%=0,76%`
