\(2KOH+CuSO_4-->K_2SO_4+Cu\left(OH\right)_2\)
0,04_____0,02__________0,02________0,02
=>\(m_{KOH}=\dfrac{40.5,6}{100}=2,24\left(g\right)=>n_{KOH}=\dfrac{2,24}{56}=0,04\left(mol\right)\)
=>\(m_{CuSO_4}=\dfrac{300.16}{100}=48\left(g\right)=>n_{CuSO_4}=\dfrac{48}{160}=0,3\left(mol\right)\)
theo pthh : \(n_{KOH}=2n_{CuSO_4}=2.0,3=0,6\left(mol\right)>0,04=>n_{KOH}hết\)
=>\(m_{kt}=0,02.98=1,96\left(g\right)=>\) ko có đáp án đúng
2KOH + CuSO4 → K2SO4 + Cu(OH)2↓
\(m_{KOH}=40\times5,6\%=2,24\left(g\right)\)
\(\Rightarrow n_{KOH}=\dfrac{2,24}{56}=0,04\left(mol\right)\)
\(m_{CuSO_4}=300\times16\%=48\left(g\right)\)
\(\Rightarrow n_{CuSO_4}=\dfrac{48}{160}=0,3\left(mol\right)\)
Theo PT: \(n_{KOH}=2n_{CuSO_4}\)
Theo bài: \(n_{KOH}=\dfrac{2}{15}n_{CuSO_4}\)
Vì \(\dfrac{2}{15}< 2\) ⇒ CuSO4 dư
Theo PT: \(n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{KOH}=\dfrac{1}{2}\times0,04=0,02\left(mol\right)\)
\(\Rightarrow m_{Cu\left(OH\right)_2}=0,02\times98=1,96\left(g\right)\)
Không có đáp án đúng
Trung hoà 20ml dd H2SO4 1M bằng dd NaOH 20%.
