Question

Answer 1

4. 

\(n_{Ca}=\dfrac{24}{40}=0.6\left(mol\right)\)

\(2Ca+O_2\underrightarrow{^{^{t^0}}}2CaO\)

\(0.6.....0.3..........0.6\)

\(V_{O_2}=0.3\cdot22.4=6.72\left(l\right)\)

\(m_{CaO}=0.6\cdot56=33.6\left(g\right)\)

5.

\(n_{C_2H_6O}=\dfrac{23}{46}=0.5\left(mol\right)\)

\(n_{O_2}=\dfrac{44.8}{22.4}=2\left(mol\right)\)

\(C_2H_6O+3O_2\underrightarrow{^{^{t^0}}}2CO_2+3H_2O\)

Lập tỉ lệ : 

\(\dfrac{0.5}{1}< \dfrac{2}{3}\Rightarrow O_2dư\)

\(m_{O_2\left(dư\right)}=\left(2-0.5\cdot3\right)\cdot32=16\left(g\right)\)

\(m_{CO_2}=0.5\cdot2\cdot44=44\left(g\right)\)

 

Answer 2

\(2KMnO_4\underrightarrow{^{^{t^0}}}K_2MnO_4+MnO_2+O_2\)

\(2KClO_3\underrightarrow{^{^{t^0}}}2KCl+3O_2\)