Đáp án:
$m_{dd\ CuSO_4}=16$ (gam)
$C\%_{dd\ sau\ phan\ ung}\approx 19,19\%$
Giải thích các bước giải:
Gọi số mol $Fe$ phản ứng là $x$ (mol)
$Fe+CuSO_4\to FeSO_4+Cu\\\; x\,\,\,\to \;\;\;x\qquad \qquad x \qquad \;\;\;\;\,\;x\;\;\; (mol)$Theo PTHH: $n_{Cu}=x$ (mol)
$\Rightarrow \Delta_{m}=m_{Cu}-m_{Fe}=5,16-5=0,16$ (gam)
$\Leftrightarrow 64x-56x=0,16$
$\Leftrightarrow x=0,02$ (mol)
Theo PTHH: $n_{CuSO_4}=x=0,02$ (mol)
$\Rightarrow m_{dd\ CuSO_4}=\dfrac{0,02\cdot 160}{20\%}=16$ (gam)
BTKL, ta có: $m_{Fe}+m_{dd\ CuSO_4}=m_{dd\ sau\ phan\ ung}+m_{Fe\ sau\ phan\ ung}$
$\Leftrightarrow m_{dd\ sau\ phan\ ung}=5+16-5,16=15,84$ (gam)
Theo PTHH: $n_{FeSO_4}=x=0,02$ (mol)
$\Rightarrow m_{FeSO_4}=152\cdot 0,02=3,04$ (gam)
$\Rightarrow C\%_{dd\ sau\ phan\ ung}=\dfrac{3,04}{15,84}\cdot 100\%\approx 19,19\%$
