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Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
a. PTHH: Fe + H2SO4 ---> FeSO4 + H2
b. Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
=> \(V_{H_2}=0,1.22,4=2,24\left(lít\right)\)
c. Theo PT: \(n_{H_2SO_4}=n_{Fe}=0,1\left(mol\right)\)
=> \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)
=> \(C_{\%_{H_2SO_4}}=\dfrac{9,8}{100}.100\%=9,8\%\)
\(n_{Al\left(OH\right)_3}=\dfrac{7,8}{78}=0,1mol\\ 2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\\ 0,1................0,15.............0,05............0,3\\ C_{\%H_2SO_4}=\dfrac{0,15.98}{300}\cdot100\%=4,9\%\\ C_{\%Al_2\left(SO_4\right)_3}=\dfrac{0,05.342}{7,8+300}\cdot100\%=5,56\%\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
2a______3a__________a_______3a (mol)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
b_______b________b______b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}27\cdot2a+24b=7,8\\3a+b=\dfrac{200\cdot19,6\%}{98}=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1\cdot24=2,4\left(g\right)\\m_{Al}=5,4\left(g\right)\\n_{Al_2\left(SO_4\right)_3}=0,1\left(mol\right)=n_{MgSO_4}\\n_{H_2}=0,4\left(mol\right)\Rightarrow m_{H_2}=0,4\cdot2=0,8\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{KL}+m_{ddH_2SO_4}-m_{H_2}=207\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1\cdot342}{207}\cdot100\%\approx16,52\%\\C\%_{MgSO_4}=\dfrac{0,1\cdot120}{207}\cdot100\%\approx5,8\%\end{matrix}\right.\)
200ml = 0,2l
\(n_{KOH}=1.0,2=0,2\left(mol\right)\)
Pt : \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O|\)
2 1 1 2
0,2 0,1
\(n_{H2SO4}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{H2SO4}=0,1.98=9,8\left(g\right)\)
\(C_{ddH2SO4}=\dfrac{9,8.100}{49}=20\)0/0
Chúc bạn học tốt
cho 3,2 g cuo tác dụng vừa đủ với dung dịch H2SO4 , 4,9%. Tính nồng độ phần trăm của dung dịch CuSO4
nCuO=0,04 mol
CuO + H2SO4 =>CuSO4 + H2O
0,04 mol=>0,04 mol=>0,04 mol
mH2SO4=0,04.98=3,92 gam
=>m dd H2SO4=3,92/4,9%=80 gam
mCuSO4 sau=0,04.160=6,4 gam
mdd CuSO4=3,2+80=83,2 gam
C% dd CuSO4=6,4/83,2.100%=7,69%
cho \(m_{CuO}=3,2g\Rightarrow n_{CuO}=\frac{3,2}{80}=0,04mol\)
PTHH:
CuO + H2SO4 -> CuSO4 + H2O
0,04mol----------->0,04mol--------->0,04mol
ta có: \(m_{H_2SO_4}=0,04.98=3,92g\)
\(C\%_{d^2H_2SO_{4_{ }}}=4,9\%\)
=. \(m_{d^2H_2SO_4}=\frac{m_{H_2SO_4}.100}{C\%}=\frac{3,92.100}{4,9}=80g\)
áp dụng ĐLBTKL ta có: \(m_{d^2CUSO_4}=m_{CuO}+m_{d^2H_2SO_4}=3,2+80=83,2g\)
\(m_{CuSO_4}=0,04.160=6,4g\)
\(\Rightarrow C\%_{d^2CuSO_4}=\frac{m_{CuSO_4}}{m_{d^2CuSO_4}}.100=\frac{6,4}{83,2}.100=7,69\%\)
\(m_{H_2SO_4}=\frac{200.19,6}{100}=39,2g\)
\(\rightarrow n_{H_2SO_4}=\frac{39,2}{98}=0,4mol\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
x x x x
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
y 1,5y 0,5y 1,5y
Đặt \(\hept{\begin{cases}x\left(mol\right)=Mg\\y\left(mol\right)=Al\end{cases}}\)
a. Có hệ phương trình là: \(\hept{\begin{cases}24x+27y=7,8\\x+1,5y=0,4\end{cases}}\)
\(\rightarrow\hept{\begin{cases}x=0,1\\y=0,2\end{cases}}\)
\(m_{Mg}=0,1.24=2,4g\)
\(m_{Al}=7,8-2,4=5,4g\)
b. \(n_{H_2}=0,1+1,5.0,2=0,4mol\)
\(\rightarrow m_{H_2}=0,4.2=0,8g\)
\(m_{ddsaupu}=m_{hh}+m_{ddH_2SO_4}-m_{H_2}\)
\(\rightarrow m_{ddsaupu}=7,8+200-0,8=207g\)
Những dung dịch thu được sau phản ứng: \(MgSO_4;Al_2\left(SO_4\right)_3\)
\(m_{MgSO_4}=0,1.\left(24+32+16.4\right)=12g\)
\(\rightarrow C\%_{MgSO_4}=\frac{12.100}{207}=5,8\%\)
\(m_{Al_2\left(SO_4\right)_3}=0,5.0,2.\left(27.2+32.3+16.12\right)=34,2g\)
\(\rightarrow C\%_{Al_2\left(SO_4\right)_3}=\frac{34,2.100}{207}=16,52\%\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + H2SO4 → FeSO4 + H2
Mol: 0,1 0,1
\(C\%_{ddH_2SO_4}=\dfrac{0,1.98.100\%}{100}=9,8\%\)