a, \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
\(m_{CuCl_2}=270.10\%=27\left(g\right)\Rightarrow n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\)
Ta có: \(\dfrac{0,15}{1}< \dfrac{0,2}{1}\) ⇒ Fe hết, CuCl2 dư
PTHH: Fe + CuCl2 ---> FeCl2 + Cu
Mol: 0,15 0,15 0,15 0,15
\(a=m_{Cu}=0,15.64=9,6\left(g\right)\)
b, \(m_{dd.sau.pứ}=8,4+270-9,6=268,8\left(g\right)\)
\(m_{CuCl_2dư}=\left(0,2-0,15\right).135=6,75\left(g\right)\)
\(\left\{{}\begin{matrix}C\%_{CuCl_2dư}=\dfrac{6,75.100\%}{268,8}=2,51\%\\C\%_{FeCl_2}=\dfrac{0,15.127.100\%}{268,8}=7,09\%\end{matrix}\right.\)
c, \(V_{ddCuCl_2}=\dfrac{270}{1,35}=200\left(ml\right)=0,2\left(l\right)\)
\(\left\{{}\begin{matrix}C_{M_{CuCl_2dư}}=\dfrac{0,2-0,15}{0,2}=0,25M\\C_{M_{FeCl_2}}=\dfrac{0,15}{0,2}=0,75M\end{matrix}\right.\)
\(Fe+CuCl_2\rightarrow Cu+FeCl_2\)
mCuCl2 = \(\dfrac{270.10}{100}=27\left(g\right)\)
\(nCuCl_2=\dfrac{27}{64+71}=0,2\left(mol\right)\)
\(nFe=\dfrac{8,4}{56}=0,15\left(mol\right)\)
CuCl dư
=> n Cu = nFe = 0, 15(mol)
=> a = mCu = 0,15 . 64 = 9,6(g)
mdd = mFe + mCuCl2 - mCu
= 8,4 + 270 - 9, 6 = 268,8(g)
mFeCl2 = 0,15 . ( 56 + 71 ) = 19,05 (g)
C%ddA = \(\dfrac{19,05.100}{268,8}=7,09\%\)
c) D = 1,35 . 10 = 13,5 g /ml
V dd A = mdd A / D = 13,725 / 13,5 = 61/60 (l)
CM = \(\dfrac{n}{V}=\dfrac{0,15}{\dfrac{61}{60}}=0,15\) mol / l