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nMg = mMg / M(Mg) = 3 / 24 = 0,125 (mol)
m\(_{H2SO4}\) = (m\(_{dd}\) * C%) / 100% = ( 150 * 1,96%) / 100% = 2,94 (gam)
nH2SO4 = m\(_{H2SO4}\) / M\(_{H2SO4}\) = 2,94 / 98 = 0,03 (mol)
a, Mg + H2SO4 → MgSO4 + H2
(mol) 0,125......0,03
xét tỉ lệ : \(\frac{0,125}{1}\) > \(\frac{0,03}{1}\)
pư : 0,03.................0,03...........0,03...........0,03
dư : (0,125-0,03)
0,095 (mol)
vậy : V(H2) = 0,03 * 22,4 = 0,672 (lít)
b, Dung dịch sau pư có MgSO4 : 0,03 (mol)
mặt khác Mg còn dư sau pư : 0,095 mol
m(dd sau) = m(Mg bđ) + m(dd H2SO4 ) - m(H2) - m(Mg dư)
= 3 + 150 - 0,03 * 2 - 0,095*24
= 150,66 (gam)
Vậy : C%(MgSO4) = {mMgSO4 / m(dd sau) } * 100%
= (0,03*120/150,66) * 100%
~ 2,39 (%)
nMg= 3/24=0.125 mol
mH2SO4= 150*1.96/100=2.94g
nH2SO4= 2.94/98=0.03 mol
Mg + H2SO4 --> MgSO4 + H2
Bđ: 0.125__0.03
Pư : 0.03___0.03______0.03____0.03
Kt: 0.095___0_________0.03___0.03
VH2= 0.03*22.4=0.672l
mdd sau phản ứng = mMg(bđ) + mddH2SO4 - mH2 - mMg(dư)
=> 3 + 150 - 0.06 - 0.095*24=150.66g
mMgSO4= 0.03*120=3.6g
C%MgSO4= 3.6/150.66*100%= 2.39%
1.
nAl=\(\dfrac{5,4}{27}\)=0,2 mol
mHCl=\(\dfrac{175.14,6}{100}\)=25,55g
nHCl=\(\dfrac{25,55}{36,5}\)=0,7
2Al + 6HCl → 2AlCl3 + 3H2↑
n trước pứ 0,2 0,7
n pứ 0,2 →0,6 → 0,2 → 0,3 mol
n sau pứ hết dư 0,1
Sau pứ HCl dư.
mHCl (dư)= 36,5.0,1=3,65g
mcác chất sau pư= 5,4 +175 - 0,3.2= 179,8g
mAlCl3= 133,5.0,2=26,7g
C%ddHCl (dư)= \(\dfrac{3,65.100}{179,8}=2,03%\)%
C%ddAlCl3 = \(\dfrac{26,7.100}{179,8}\)= 14,85%
2.
200ml= 0,2l
mMg= \(\dfrac{4,2}{24}=0,175mol\)
Mg + 2HCl → MgCl2 + H2↑
0,175→ 0,35 → 0,175→0,175 mol
a) VH2= 0,175.22,4=3,92l.
b)C%dHCl= \(\dfrac{0,35}{0,2}=1,75\)M
n H2SO4=\(\dfrac{10\%.490}{2+32+16.4}=0,5mol\)
n Al2O3 =\(\dfrac{10,2}{27.2+16.3}=0,1mol\)
\(Al_2O_3+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2O\)
bđ 0,1............0,5
pư 0,1............0,3..................0,1
spu 0 ................0,2................0,1
=> sau pư gồm H2SO4 dư , Al2(S04)3 và H2O
m H2SO4 dư = \(0,2.\left(2+32+16.3\right)=19,6g\)
m Al2(SO4)3 = \(0,1\left(27.2+32.3+16.4.3\right)=34,2g\)
m dd = \(490+10,2=500,2g\)
% Al2(SO4)3 = \(\dfrac{34,2}{500,2}.100\sim6,84\%\)
% H2SO4 dư = \(\dfrac{19,6}{500,2}.100\sim3,92\%\)
Mg+2CH3COOH->(CH3COO)2Mg+H2
0,15------0,3-------------0,15-------------0,15
n Mg=\(\dfrac{3,6}{24}\)=0,15 mol
m CH3COOH=24g =>n CH3COOH=\(\dfrac{24}{60}\)=0,6 mol
->CH3COOH dư
=>C% (CH3COO)2Mg=\(\dfrac{0,15.142}{200+3,6-0,15.2}\).100=10,48%
=>C% CH3COOH dư= \(\dfrac{0,3.60}{200+3,6-0,15.2}\).100=8,85%
a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(m_{H_2SO_4}=147.10\%=14,7\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)
PTHH: Zn + H2SO4 → ZnSO4 + H2
Mol: 0,1 0,1 0,1 0,1
Ta có: \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\) ⇒ Zn hết, H2SO4 dư
b) \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c) mdd sau pứ = 6,5 + 147 - 0,1.2 = 153,3 (g)
\(C\%_{ddZnSO_4}=\dfrac{0,1.161.100\%}{153,3}=10,502\%\)
\(C\%_{ddH_2SO_4dư}=\dfrac{\left(0,15-0,1\right).98.100\%}{153,3}=3,196\%\)
PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\\n_{H_2SO_4}=\dfrac{784\cdot10\%}{98}=0,8\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,6}{1}< \dfrac{0,8}{1}\) \(\Rightarrow\) H2SO4 còn dư, Fe phản ứng hết
\(\Rightarrow\left\{{}\begin{matrix}n_{FeSO_4}=n_{H_2}=0,6mol\\n_{H_2SO_4\left(dư\right)}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,6\cdot152=91,2\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,2\cdot98=19,6\left(g\right)\\m_{H_2}=0,6\cdot2=1,2\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Fe}+m_{ddH_2SO_4}-m_{H_2}=816,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeSO_4}=\dfrac{91,2}{816,4}\cdot100\%\approx11,17\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{19,6}{816,4}\cdot100\%\approx2,4\%\end{matrix}\right.\)
\(n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\)
\(n_{H2SO4}=\dfrac{784.10\%}{98}=0,8\left(mol\right)\)
PTHH : \(Fe+H_2SO_4-->FeSO_4+H_2\uparrow\)
Theo pthh : \(n_{H2}=n_{FeSO4}=n_{H2SO4\left(pứ\right)}=n_{Fe}=0,6\left(mol\right)\)
\(\Rightarrow n_{H2SO4\left(dư\right)}=0,8-0,6=0,2\left(mol\right)\)
Áp dụng ĐLBTKL :
mFe + m(dd H2SO4) = m(ddspu) + mH2
=> 33,6 + 784 = m(ddspu) + 0,6.2
=> m(ddspu) = 816,4(g)
\(\Rightarrow\left\{{}\begin{matrix}C\%FeSO_{\text{4}}=\dfrac{0,6.152}{816,4}\cdot100\%\approx11,17\%\\C\%H_2SO_{4\left(dư\right)}=\dfrac{0,2.98}{816,4}\cdot100\%\approx2,4\%\end{matrix}\right.\)