a)
$Cu(OH)_2 \xrightarrow{t^o} CuO + H_2O$
$2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O$
Vì theo bảo toàn khối lượng :
$m_{hh} = m_{oxit} + m_{H_2O}$
mà $m_{H_2O} > 0$ nên $m_{hh} > m_{oxit}$
Do đó khối lượng rắn giảm.
b)
Gọi $n_{Cu(OH)_2} = a ; n_{Fe(OH)_3} = b$
$\Rightarrow 98a + 107b = 6,06(1)$
Theo PTHH :
$m_{cr} = 80a + 80b = 4,8(2)$
Từ (1)(2) suy ra a = 0,04; b = 0,02
Suy ra :
$\%m_{Cu(OH)_2} = \dfrac{0,04.98}{6,06}.100\% = 64,67\%$
$\%m_{Fe(OH)_3} = 35,33\%$
\(n_{Cu\left(OH\right)_2}=a\left(mol\right),n_{Fe\left(OH\right)_3}=b\left(mol\right)\)
\(m_{hh}=98a+107b=6.06\left(g\right)\left(1\right)\)
\(Cu\left(OH\right)_2\underrightarrow{^{^{t^0}}}CuO+H_2O\)
\(a............a\)
\(2Fe\left(OH\right)_3\underrightarrow{^{^{t^0}}}Fe_2O_3+3H_2O\)
\(b............0.5b\)
\(m_{Cr}=80a+0.5b\cdot160=4.8\left(g\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.04,b=0.02\)
\(\%Cu\left(OH\right)_2=\dfrac{0.04\cdot98}{6.06}\cdot100\%=64.68\%\)
\(\%Fe\left(OH\right)_3=35.32\%\)