2NaOH + CuCl2-------> 2NaCl + Cu(OH)2
(mol) 2 1 2 1
(mol) 0,5 0,25 0,5 0,25
khối lượng NaOH:
C%=\(\dfrac{m_{ct}}{m_{dd}}\) .100⇒ mct=\(\dfrac{C\%.m_{dd}}{100}\)=\(\dfrac{10.200}{100}\)=20g
số mol NaOH:
n=\(\dfrac{m}{M}\)=\(\dfrac{20}{40}\)=0,5(mol)
khối lượng Cu(OH)2:
m=n.M=0,25.98=24,5g
CHỖ NÀO HỔNG HỈU HỎI MÌNH NHA
a) 2NaOH + CuCl2 → 2NaCl + Cu(OH)2↓
b) \(m_{NaOH}=200\times10\%=20\left(g\right)\)
\(\Rightarrow n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
Theo PT: \(n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\times0,5=0,25\left(mol\right)\)
\(\Rightarrow m_{Cu\left(OH\right)_2}=0,25\times98=24,5\left(g\right)\)
