Question

Answer 1

a)

Gọi số mol CH₄, C₄H₁₀ là a, b (mol)

\(\Rightarrow\left\{{}\begin{matrix}a+b=\dfrac{7,84}{22,4}=0,35\\\overline{M}=\dfrac{16a+58b}{a+b}=23.2=46\left(g/mol\right)\end{matrix}\right.\)

=> a = 0,1 (mol); b = 0,25 (mol)

PTHH: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

              0,1--->0,2

            \(2C_4H_{10}+13O_2\underrightarrow{t^o}8CO_2+10H_2O\)

              0,25---->1,625

=> \(V_{O_2}=22,4.\left(0,2+1,625\right)=40,88\left(l\right)\)

b) \(n_{O_2}=1,825\left(mol\right)\)

PTHH: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

               3,65<-------------------------1,825

\(\Rightarrow m_{KMnO_4\left(LT\right)}=3,65.158=576,7\left(g\right)\)

\(\Rightarrow m_{KMnO_4\left(TT\right)}=\dfrac{576,7.100}{80}=720,875\left(g\right)\)