1. C

2. D

3. C

4. A

5. D

6. C

7. D

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Bảo toàn nguyên tố C:

\(n_{CaCO_3}=n_{CO_2}=0,25\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=0,25.100=25\left(g\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

\(\rightarrow nNaOH=\dfrac{8}{40}=0,2mol\rightarrow nH_2SO_4=0.1mol\)

\(\rightarrow VdungdichH_2SO_4=\dfrac{0,1}{1}=0,1l\)

\(n_{H_2SO_4}=\dfrac{1}{2}n_{H_+}=\dfrac{1}{2}n_{OH^-}=\dfrac{0,2}{2}=0,1\left(mol\right)\)

\(\Rightarrow V=0,1\left(l\right)\)

\(n_{NaOH}=\dfrac{8}{40}=0,2(mol)\\ PTHH:NaOH+HCl\to NaCl+H_2O\\ \Rightarrow n_{HCl}=n_{NaOH}=0,2(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,2}{1}=0,2(l)\)

nó hiểu con quá mà đứa nào cx 1 điểm, 15 câu trl đúng 1 câu....

\(n_{Al}=\dfrac{2,7}{27}=0,1(mol)\\ PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow n_{HCl}=3n_{Al}=0,3(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,3}{0,5}=0,6(l)\)

Bảo toàn nguyên tố Cl, Al:

\(n_{HCl}=n_{Cl}=3n_{AlCl_3}=3n_{Al}=0,3\left(mol\right)\)

\(\Rightarrow V=\dfrac{0,3}{0,5}=0,6\left(l\right)\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ PTHH:H_2+2O_2\xrightarrow{t^o}2H_2O\\ \Rightarrow n_{H_2O}=2n_{H_2}=0,2(mol)\\ \Rightarrow m_{H_2O}=0,2.18=3,6(g)\)

Bảo toàn nguyên tố H:

\(n_{H_2O}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{H_2O}=1,8\left(g\right)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1(mol)\\ PTHH:Zn+H_2SO_4\to ZnSO_4+H_2\\ \Rightarrow n_{ZnSO_4}=n_{Zn}=0,1(mol)\\ \Rightarrow m_{ZnSO_4}=0,1.161=16,1(g)\)

Bảo toàn nguyên tố Zn:

\(n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow m_{ZnSO_4}=16,1\left(g\right)\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{Fe}=0,2(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,2}{0,5}=0,4(l)\)

Bảo toàn nguyên tố Cl, Fe:

\(n_{HCl}=2n_{FeCl_2}=2n_{Fe}=0,2\left(mol\right)\)

\(\Rightarrow V=0,4\left(l\right)\)