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a) \(m_{dd}=\dfrac{200.100}{10}=2000\left(g\right)\)
b) mH2O = 2000 - 200 = 1800 (g)
\(a.m_{ddNaCl}=\dfrac{15}{5}\cdot100=300g\\ b.m_{nước}+m_{muối}=m_{dd,muối}\\ \Rightarrow m_{nước}=m_{dd,muối}-m_{muối}\\ =300-15\\ =285g\)
$Fe+2HCl\rightarrow FeCl_2+H_2$
$n_{Fe}=\dfrac{11,2}{56}=0,2\ mol$
a) Khối lượng dung dịch HCl tối thiểu cần dùng:
Theo phương trình:
$n_{HCl}=2n_{Fe}=0,4\ mol$
$m_{HCl}=0,4\cdot36,5=14,6\ gam$
$m_{dd\ HCl}=\dfrac{14,6\cdot100}{7,3}=200\ gam$
b) Khối lượng muối tạo thành:
$n_{FeCl_2}=n_{Fe}=0,2\ mol$
$m_{FeCl_2}=0,2\cdot127=25,4\ gam$
c) Nồng độ phần trăm dung dịch muối khi axit vừa đủ:
$n_{H_2}=n_{Fe}=0,2\ mol$
$m_{H_2}=0,2\cdot2=0,4\ gam$
$m_{dd\ sau\ phản\ ứng}=200+11,2-0,4=210,8\ gam$
$C\%_{FeCl_2}=\dfrac{25,4}{210,8}\cdot100\%\approx12,05\%$
d) Cho $11,2$ gam Fe vào $300$ gam dung dịch HCl $7,3\%$:
$m_{HCl}=\dfrac{300\cdot7,3}{100}=21,9\ gam$
$n_{HCl}=\dfrac{21,9}{36,5}=0,6\ mol$
Theo phương trình, $n_{HCl\ phản\ ứng}=0,4\ mol$.
Vì $0,6>0,4$ nên HCl dư.
$n_{HCl\ dư}=0,6-0,4=0,2\ mol$
$m_{HCl\ dư}=0,2\cdot36,5=7,3\ gam$
$m_{FeCl_2}=25,4\ gam$
$m_{dd\ sau\ phản\ ứng}=300+11,2-0,4=310,8\ gam$
Dung dịch sau phản ứng gồm $FeCl_2$ và $HCl$ dư.
$C\%_{FeCl_2}=\dfrac{25,4}{310,8}\cdot100\%\approx8,17\%$
$C\%_{HCl}=\dfrac{7,3}{310,8}\cdot100\%\approx2,35\%$
Vậy:
$a)\ m_{dd\ HCl}=200\ gam$
$b)\ m_{FeCl_2}=25,4\ gam$
$c)\ C\%_{FeCl_2}\approx12,05\%$
$d)\ C\%_{FeCl_2}\approx8,17\%;\quad C\%_{HCl}\approx2,35\%$
a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, \(n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)
c, m dd muối = 13,6 + 172,8 = 186,4 (g)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{13,6}{186,4}.100\%\approx7,3\%\)
\(pthh:Zn+2HCl--->ZnCl_2+H_2\uparrow\)
a. Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
Theo pt: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(lít\right)\)
b. Theo pt: \(n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)
c. \(C_{\%_{ZnCl_2}}=\dfrac{m_{ZnCl_2}}{m_{dd_{ZnCl_2}}}.100\%=\dfrac{13,6}{13,6+172,8}.100\%=7,3\%\)
a, nFe = 0,56/56 = 0,01 (mol)
PTHH: Fe + H2SO4 -> FeSO4 + H2
Mol: 0,01 ---> 0,01 ---> 0,01 ---> 0,01
mFeSO4 = 0,01 . 152 = 1,52 (g)
VH2 = 22,4 . 0,01 = 0,224 (l)
b, mH2SO4 = 0,01 . 98 = 0,98 (g)
c, mddH2SO4 = 0,98/19,6% = 5 (g)
d, mdd (sau p/ư) = 5 + 0,56 = 5,56 (g)
C%FeSO4 = 1,52/5,56 = 27,33%
a, nFe = 0,56/56 = 0,01 (mol)
PTHH: Fe + H2SO4 -> FeSO4 + H2
Mol: 0,01 ---> 0,01 ---> 0,01 ---> 0,01
mFeSO4 = 0,01 . 152 = 1,52 (g)
VH2 = 22,4 . 0,01 = 0,224 (l)
b, mH2SO4 = 0,01 . 98 = 0,98 (g)
c, mddH2SO4 = 0,98/19,6% = 5 (g)
d, mdd (sau p/ư) = 5 + 0,56 = 5,56 (g)
C%FeSO4 = 1,52/5,56 = 27,33%
\(n_{HCl}=1\cdot0,2=0,2\left(mol\right)\\ PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\\ a,n_{MgO}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\\ \Rightarrow m=m_{MgO}=0,1\cdot40=4\left(g\right)\\ b,n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\\ \Rightarrow m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\\ c,m_{CT_{HCl}}=0,2\cdot36,5=7,3\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{7,3}{250}\cdot100\%=2,92\%\)
\(n_{H_2O}=n_{MgO}=0,1\left(mol\right)\\ \Rightarrow m_{H_2O}=0,1\cdot18=1,8\left(g\right)\\ \Rightarrow m_{dd_{MgCl_2}}=4+250-1,8=252,2\left(g\right)\\ \Rightarrow C\%_{MgCl_2}=\dfrac{9,5}{252,2}\cdot100\%\approx3,77\%\)
\(n_{Na_2CO_3}=\dfrac{71,5}{286}=0,25\left(mol\right)\\ \Rightarrow m_{Na_2CO_3}=0,25.106=26,5\left(g\right)\\m_{H_2O\left(thêm\right)}=x\left(g\right)\Rightarrow V_{H_2O\left(thêm\right)}=x\left(ml\right)\\ C\%_{ddNa_2CO_3\left(sau\right)}=8\%\\ \Leftrightarrow\dfrac{26,5}{71,5+x}.100\%=8\%\\ \Leftrightarrow x=259,75\left(ml\right)\)
\(m_{NaCl}=200.8\%=16\left(g\right)\\ \rightarrow m_{ddmuối\left(5\%\right)}=\dfrac{16}{5\%}=320\left(g\right)\\ \rightarrow m_{H_2O\left(thêm\right)}=320-200=120\left(g\right)\)