Ta có: \(m_{CuSO_4}=\dfrac{10\%.160}{100\%}=16\left(g\right)\)
=> \(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)
Ta lại có: \(m_{NaOH}=\dfrac{5\%.240}{100\%}=12\left(g\right)\)
=> \(n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\)
a. PTHH: \(CuSO_4+2NaOH--->Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\)
Vậy NaOH dư, CuSO4 hết.
Theo PT: \(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,1\left(mol\right)\)
=> \(m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\)
b. Ta có: \(m_{dd_{Na_2SO_4}}=240+16-9,8=246,2\left(g\right)\)
Ta có: \(m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)
=> \(C_{\%_{Na_2SO_4}}=\dfrac{14,2}{246,2}.100\%=5,77\%\)
Tiếp:
Theo PT: \(n_{NaOH}=2.n_{CuSO_4}=2.0,1=0,2\left(mol\right)\)
=> \(m_{NaOH_{PỨ}}=0,2.40=8\left(g\right)\)
=> \(m_{NaOH_{dư}}=12-8=4\left(g\right)\)
=> \(C_{\%_{NaOH_{dư}}}=\dfrac{4}{240}.100\%=1,7\%\)
\(a,m_{CuSO_4}=160.10\%=16\left(g\right)\\ n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\\ b,m_{NaOH}=240.5\%=12\left(g\right)\\ n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\\ PTHH:2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\\ Vì:\dfrac{n_{NaOH\left(đề\right)}}{n_{NaOH\left(PTHH\right)}}=\dfrac{0,3}{2}>\dfrac{n_{CuSO_4\left(đề\right)}}{n_{CuSO_4\left(PTHH\right)}}=\dfrac{0,1}{1}\\ \Rightarrow CuSO_4hết,NaOHdư\)
Vậy:
\(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,1\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\)
\(b,m_{ddB}=m_{ddCuSO_4}+m_{ddNaOH}-m_{\downarrow}=160+240-9,8=390,2\left(g\right)\)
\(n_{NaOH\left(dư\right)}=0,3-0,1.2=0,1\left(mol\right)\)
\(C\%_{ddNa_2SO_4}=\dfrac{0,1.142}{390,2}.100\approx3,639\%\\ C\%_{ddNaOH\left(dư\right)}=\dfrac{0,1.40}{390,2}.100\approx1,025\%\)
