a) \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

\(n_{Al}=\dfrac{m}{M}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

Theo PTHH, \(n_{Al_2O_3}=\dfrac{2}{4}\cdot n_{Al}=0,5\cdot0,1=0,05\left(mol\right)\)

\(m_{Al_2O_3}=n\cdot M=0,05\cdot102=5,1\left(g\right)\)

b) Theo PTHH, \(n_{O_2}=\dfrac{3}{4}\cdot n_{Al}=0,75\cdot0,1=0,075\left(mol\right)\)

\(V_{O_2}=n\cdot22,4=0,075\cdot22,4=1,68\left(l\right)\)

\(V_{kk}=5\cdot V_{O_2}=5\cdot1,68=8,4\left(l\right)\)

c) \(2KMnO_4\underrightarrow{t^o}MnO_2+O_2\uparrow\ + K_2MnO_4\)

Theo PTHH, \(n_{KMnO_4}=2\cdot n_{O_2}=2\cdot0,075=0,15\left(mol\right)\)

\(m_{KMnO_4}=n\cdot M=0,15\cdot155=23,25\left(g\right)\)

Bài 2:

a) \(Mg+H_2SO_4\rightarrow\ MgSO_4+H_2\uparrow\)

\(n_{Mg}=\dfrac{m}{M}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

Theo PTHH, \(n_{H_2}=n_{Mg}=0,1\left(mol\right)\)

\(V_{H_2}=n\cdot22,4=0,1\cdot22,4=2,24\left(l\right)\)

b) Theo PTHH, \(n_{MgSO_4}=n_{Mg}=0,1\left(mol\right)\)

\(m_{MgSO_4}=n\cdot M=0,1\cdot120=12\left(g\right)\)

c) Theo PTHH, \(n_{H_2SO_4}=n_{Mg}=0,1\left(mol\right)\)

\(m_{H_2SO_4}=n\cdot M=0,1\cdot98=9,8\left(g\right)\)

Bài 3:

a) \(2Na+2H_2O\rightarrow\ 2NaOH+H_2\uparrow\)

b) \(n_{Na}=\dfrac{m}{M}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

Theo PTHH, \(n_{H_2}=\dfrac{1}{2}\cdot n_{Na}=0,5\cdot0,2=0,1\left(mol\right)\)

\(V_{H_2}=n\cdot 22,4=0,1\cdot22,4=2,24\left(l\right)\)

c) \(m_{dd}=m_{ct}+m_{dm}=4,6+150=154,6\left(g\right)\)

⇒ \(C\%=\dfrac{m_{ct}.100\%}{m_{dd}}=\dfrac{4,6.100}{154,6}\approx\ 3\left(\%\right)\)

Đáp án:

Giải thích các bước giải: