1: CaCl₂ + 2AgNO₃ ---> 2AgCl + Ca(NO₃)₂

2: 2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

3: P₂O₅   +  3H₂O --->  2H₃PO₄

4: 4FeO + O₂ ---> 2Fe₂O₃

Câu 2: 

M_CaCO3 = 100(g/mol)

%Ca = \(\dfrac{40}{100}.100\%\)= 40%

%C = \(\dfrac{12}{100}.100\)% = 12%

%O = \(\dfrac{3.16}{100}\).100% = 48%

Câu 3:

nCO₂ = \(\dfrac{5,6}{22,4}\)= 0,25 mol => mCO₂ = 0,25.44 = 11 gam

nCO₂ = \(\dfrac{9.10^{23}}{6,022.10^{23}}\)≃ 1,5 mol => mCO₂ = 1,5. 44 = 66 gam

Câu 4: 

2Al  +  6HCl  -->  2AlCl₃  +   3H₂ 

nAl = 2,7/27 = 0,1 mol. Theo tỉ lệ phản ứng => nAlCl₃ = nAl = 0,1 mol

=> mAlCl₃ = 0,1.133,5 = 13,35 gam

Đáp án A

Đáp án A.

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: Fe + H₂SO₄ ---> FeSO₄ + H₂

             0,1                                     0,1

\(m_{Fe}=0,1.56=5,6\left(g\right)\\ \%m_{Fe}=\dfrac{5,6}{12,8}=43,75\%\\ \%m_{FeO}=100\%-43,75\%=56,25\%\)

thành phần phấn trăm của từng chất à

\(M_{\left(NH_4\right)_2SO_4}=\left(14+1.4\right).2+32+16.4=132\left(\dfrac{g}{mol}\right)\)

\(\%N=\dfrac{14.2}{132}.100\%=21,2\%\)

\(\%H=\dfrac{1.4.2}{132}.100\%=6\%\)

\(\%S=\dfrac{32}{132}.100\%=24,2\%\)

\(\%O=\dfrac{16.4}{132}.100\%=48,5\%\)

\(n_{FeO}=a\left(mol\right),n_{ZnO}=b\left(mol\right)\)

\(m_{hh}=72a+81b=15.3\left(g\right)\left(1\right)\)

\(TC:n_{FeO\left(pư\right)}=0.8a\left(mol\right),n_{ZnO\left(pư\right)}=0.8b\left(mol\right)\)

\(FeO+CO\underrightarrow{t^0}Fe+CO_2\)

\(ZnO+CO\underrightarrow{t^0}Zn+CO_2\)

\(BTKL:\)

\(15.3+28n_{CO}=12.74+44n_{CO}\)

\(\Rightarrow n_{CO}=\dfrac{15.3-12.74}{44-28}=0.16\left(mol\right)\)

\(\Rightarrow0.8a+0.8b=0.16\left(2\right)\)

\(\left(1\right),\left(2\right):a=b=0.1\)

\(\%FeO=\dfrac{0.1\cdot72}{15.3}\cdot100\%=47.06\%\)

\(\%ZnO=52.94\%\)