Đáp án:
b,
`-` `V_{H_2\ \text{(đktc)}}=5,6\ (l).`
c,
`-` `m_{Cu(OH)_2}=12,25\ (g).`
d,
`-` `C%_{NaOH\ \text{(dư)}}=2%`
`-` `C%_{Na_2SO_4}=3,56%`
Giải thích các bước giải:
`-` `n_{Na}=\frac{11,5}{23}=0,5\ (mol).`
`-` `n_{CuSO_4}=\frac{500\times 4%}{160}=0,125\ (mol).`
a,
Phương trình hóa học:
`2Na + 2H_2O \to 2NaOH + H_2\uparrow\ (1)`
`2NaOH + CuSO_4 \to Na_2SO_4 + Cu(OH)_2\downarrow\ (2)`
b,
`-` Theo phương trình `(1):` `n_{H_2}=\frac{1}{2}n_{Na}=0,25\ (mol).`
`\to V_{H_2\ \text{(đktc)}}=0,25\times 22,4=5,6\ (l).`
c,
`-` Tỉ lệ: `n_{NaOH}:n_{CuSO_4}=\frac{0,5}{2}>\frac{0,125}{1}`
`\to NaOH` dư, `CuSO_4` hết.
`-` Theo phương trình `(2):` `n_{Cu(OH)_2}=n_{CuSO_4}=0,125\ (mol).`
`\to m_{Cu(OH)_2}=0,125\times 98=12,25\ (g).`
d,
`-` `n_{NaOH\ \text{(dư)}}=0,5-0,125\times 2=0,25\ (mol).`
`-` Theo phương trình `(2):` `n_{Na_2SO_4}=n_{CuSO_4}=0,125\ (mol).`
`-` Ta có:
`m_{\text{dd spư}}=m_{Na}+m_{dd\ CuSO_4}-m_{H_2}-m_{Cu(OH)_2}`
`\to m_{\text{dd spư}}=11,5+500-0,25\times 2-12,25=498,75\ (g).`
`\to C%_{NaOH\ \text{(dư)}}=\frac{0,25\times 40}{498,75}\times 100%=2%`
`\to C%_{Na_2SO_4}=\frac{0,125\times 142}{498,75}\times 100%=3,56%`
