\(m_{NaOH}=\dfrac{20\cdot20\%}{100\%}=4\left(g\right)\)
\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)
PTHH: \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)
Mol (pt): 2 1 1 1
Mol (đề): 0,1
a) Theo pt, ta có: \(n_{CuSO_4}=2\cdot n_{NaOH}=2\cdot0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{CuSO_4}=0,2\cdot160=32\left(g\right)\)
\(\Rightarrow m_{\text{dd }CuSO_4}=\dfrac{32\cdot100\%}{10\%}=320\left(g\right)\)
b) Theo pt, ta có: \(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,2\left(mol\right)\)
\(\Rightarrow m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\)
c) \(m_{Na_2SO_4}=0,2\cdot142=28,4\left(g\right)\)
Theo ĐLBTKL, ta có:
\(m_{\text{dd }Na_2SO_4}=m_{\text{dd }NaOH}+m_{\text{dd }CuSO_4}-m_{Cu\left(OH\right)_2}\)
\(=20+320-19,6=320,4\left(g\right)\)
\(\Rightarrow C\%_{\text{dd }Na_2SO_4}=\dfrac{28,4}{320,4}\cdot100\%\approx8,86\%\)
\(\text{#}Toru\)
