Đáp án:
`m=m_{Ag}=43,2\ (g).`
`C\%_{Cu(NO_3)_2}=8\%`
`C\%_{AgNO_3\ \text{(dư)}}=3,62\%`
Giải thích các bước giải:
`-` `n_{Cu}=\frac{12,8}{64}=0,2\ (mol)`
`m_{AgNO_3}=500\times 17\%=85\ (g)`
`\to n_{AgNO_3}=\frac{85}{170}=0,5\ (mol)`
Phương trình hóa học:
`Cu+2AgNO_3\to Cu(NO_3)_2+2Ag\downarrow`
`0,2\ \to \ \ \ 0,4\ \ \to \ \ \ \ \ 0,2\ \ \ \to 0,4\ \ \ \ (mol)`
Tỉ lệ: `n_{Cu}:n_{AgNO_3}=\frac{0,2}{1}<\frac{0,5}{2}`
`\to AgNO_3` dư.
`\to n_{AgNO_3\ \text{(dư)}}=0,5-(0,2\times 2)=0,1\ (mol).`
`\to m=m_{Ag}=0,4\times 108=43,2\ (g).`
`m_{\text{dd spư}}=m_{Cu}+m_{dd\ AgNO_3}-m_{Ag}`
`\to m_{\text{dd spư}}=12,8+500-43,2=469,6\ (g).`
`\to C\%_{Cu(NO_3)_2}=\frac{0,2\times 188}{469,6}\times 100\%=8\%`
`\to C\%_{AgNO_3\ \text{(dư)}}=\frac{0,1\times 170}{469,6}\times 100\%=3,62\%`
