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a)
$MCO_3 + 2HCl \to MCl_2 + CO_2 + H_2O$
Theo PTHH :
$n_{H_2O} = n_{CO_2} = a(mol)$
$n_{HCl} = 2n_{CO_2} = 2a(mol)$
Bảo toàn khối lượng :
$10 + 2a.36,5 = 44a + 18a + 11,1 \Rightarrow a = 0,1$
$V = 0,1.22,4 = 2,24(lít)$
b) $n_{HCl} = 2a = 0,2(mol)$
$m = \dfrac{0,2.36,5}{3,65\%} = 200(gam)$
c)
$m_{dd} = 10 + 200 - 0,1.44 = 205,6(gam)$
$C\%_{muối} = \dfrac{11,1}{205,6}.100\% = 5,4\%$
d)
$M_{MCO_3} = \dfrac{10}{0,1} = 100 \Rightarrow M = 40(Canxi)$
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(PTHH:Fe+2HCl--->FeCl_2+H_2\)
a. Theo PT: \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=0,1.127=12,7\left(g\right)\)
b. Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(lít\right)\)
\(CH_3COOH+NaCl\rightarrow CH_3COONa+HCl\)
\(2CH_3COOH+CaCO_3\rightarrow\left(CH_3COO\right)_2Ca+H_2O+CO_2\)
2 1 1 1 1 (mol)
0,08 0,04 0,04 0,04 0,04 (mol)
\(nCO_2=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)
\(mCaCO_3=0,04.100=4\left(g\right)\)
=> \(mNaCl=12,5-4=8,5\left(g\right)\)
( không thấy hh B )
c ) .
\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
1 2 1 1 (mol)
0,04 0,08 0,04 0,04 (mol)
\(mNa_2CO_3=0,04.106=4,24\left(g\right)\)
\(mNa_2CO_{3\left(thựctế\right)}=\)\(\dfrac{4,24.85\%}{100\%}=3,604\left(g\right)\)
\(n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(0.05.......0.05......0.05...........0.05\)
\(m_{Zn}=0.05\cdot65=3.25\left(g\right)\)
\(C_{M_{H_2SO_4}}=\dfrac{0.05}{0.2}=0.25\left(M\right)\)
\(m_{ZnSO_4}=0.05\cdot161=8.05\left(g\right)\)
Bài1
a) Fe+ 2HCl---.FeCl2 +H2
n\(_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\)
Theo pthh
n\(_{FeCl2}=n_{Fe}=0,2\left(mol\right)\)
m=m\(_{FeCl2}=0,2.127=25,4\left(g\right)\)
Theo pthh
n\(_{H2}=n_{Fe}=0,2\left(mol\right)\)
V=V\(_{H2}=0,2.22,4=4,48\left(l\right)\)
b)HCl+NaOH--->NaCl+H2O
Theo pthh 1
n\(_{HCl}=2n_{Fe}=0,4\left(mol\right)\)
n\(_{NaOH}=0,3.1=0,3\left(mol\right)\)
=> HCl dư
Theo pthh2
n\(_{NaCl}=n_{NaOH}=0,3\left(mol\right)\)
m\(_{NaCl}=0,3.94=28,2\left(g\right)\)
Bài 2
a) Mg+H2SO4--->MgSO4+H2
n\(_{Mg}=\frac{3,6}{24}=0,15\left(mol\right)\)
Theo pthh
n\(_{MgSO4}=n_{Mg}=0,15\left(mol\right)\)
m=m\(_{MgSO4}=0,15.120=18\left(g\right)\)
n\(_{H2}=n_{Mg}=0,15\left(mol\right)\)
V=V\(_{H2}=0,15.22,4=3,36\left(l\right)\)
b) H2SO4+ 2KOH--->K2SO4+2H2O
n\(_{H2SO4}=n_{Mg}=0,15\left(mol\right)\)
n\(_{KOH}=0,2.1=0,2\left(mol\right)\)
=> H2SO4(Do lập tỉ số)
Theo pthh
n\(_{K2SO4}=\frac{1}{2}n_{KOH}=0,1\left(mol\right)\)
m\(_{muối}=m_{K2SO4}=174.0,1=17,4\left(g\right)\)