Ta có: \(C_{\%_{CuCl_2}}=\dfrac{m_{CuCl_2}}{200}.100\%=6,75\%\)
=> \(m_{CuCl_2}=13,5\left(g\right)\)
=> \(n_{CuCl_2}=\dfrac{13,5}{135}=0,1\left(mol\right)\)
a. PTHH: \(CuCl_2+2KOH--->Cu\left(OH\right)_2\downarrow+2KCl\) (1)
Theo PT(1): \(n_{KOH}=2.n_{CuCl_2}=2.0,1=0,2\left(mol\right)\)
=> \(m_{KOH}=0,2.56=11,2\left(g\right)\)
Ta có: \(C_{\%_{KOH}}=\dfrac{11,2}{m_{dd_{KOH}}}.100\%=5,6\%\)
=> \(m_{dd_{KOH}}=200\left(g\right)\)
b. Theo PT(1): \(n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,1\left(mol\right)\)
=> \(m_{dd_{KCl}}=11,2+200-0,1.98=201,4\left(g\right)\)
Theo PT: \(n_{KCl}=n_{KOH}=0,2\left(mol\right)\)
=> \(m_{KCl}=0,2.74,5=14,9\left(g\right)\)
=> \(C_{\%_{KCl}}=\dfrac{14,9}{201,4}.100\%=7,4\%\)
c. PTHH: \(Cu\left(OH\right)_2\overset{t^o}{--->}CuO+H_2O\) (2)
Theo PT(2): \(n_{CuO}=n_{Cu\left(OH\right)_2}=0,1\left(mol\right)\)
=> \(m_{CuO}=0,1.80=8\left(g\right)\)
d. Ta có: \(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Ta có: \(T=\dfrac{n_{KOH}}{n_{CO_2}}=\dfrac{0,2}{0,3}=0,\left(6\right)< 1\)
Vậy ta có PTHH: KOH + CO2 ---> KHCO3 (CO2 dư) (3)
Theo PT(3): \(n_{KHCO_3}=n_{KOH}=0,2\left(mol\right)\)
=> \(m_{KHCO_3}=0,2.100=20\left(g\right)\)
