a. mdd X= 6,2+193,8= 200(g) ⇒C%X=\(\dfrac{6,2}{200}.100=31\%\)
b.\(n_{NaOH}=\dfrac{6,2}{40}=0,155\left(mol\right)\)
n\(_{CuSO_4}=\dfrac{200.16}{100.160}=0,2\left(mol\right)\)
PTHH: 2NaOH + CuSO4 → Na2SO4 + Cu(OH)2
_____0,155______0,0775___0,0775_____0,0775 (mol)
Ta có tỉ lệ:\(\dfrac{n_{NaOH}}{2}:\dfrac{n_{CuSO_4}}{1}=\dfrac{0,155}{2}< \dfrac{0,2}{1}\)
⇒NaOH pư hết, CuSO4 pư dư.
\(m_{Na_2SO_4}=0,0775.142=11,005\left(g\right)\)
\(m_{Cu\left(OH\right)_2}=0,775.98=7,595\left(g\right)\)
\(\Rightarrow m_{dd}saupư=200+200-7,595=392,405\left(g\right)\)
\(\Rightarrow C\%_A=\dfrac{11,005}{392,405}.100=2,8\%\)
c. PTHH: Cu(OH)2 \(\underrightarrow{t^o}\) CuO + H2O
n\(_{Cu\left(OH\right)_2}\)=0,0775(mol)⇒ nCuO=0,0775(mol)
PTHH: CuO + 2HCl→CuCl2 + H2O
0,0775_0,155 (mol)
nHCl=0,155(mol)⇒mHCl=0,155.36,5=5,6575(g)