Đáp án:

 a. \(\%V_{O_2}=32,61\%\\ \%m_{O_2}=33,241\%\)

b. \(\%V_{O_2}=33,33\%\\ \%m_{O_2}=33,33\%\)

c. \(\%V_{O_2}=33,33\%\\ \%m_{O_2}=33,33\%\)

Giải thích các bước giải:

 a. \(\%V_{O_2}=\dfrac{0,75}{0,75+1,25+0,3}\cdot 100\%=32,61\%\\ \%m_{O_2}=\dfrac{0,75\cdot 32}{0,75\cdot 32+1,25\cdot 28+0,3\cdot 44}\cdot 100\%=33,241\%\)

b. \(n_{CO_2}=\dfrac{3\cdot 10^{23}}{6\cdot 10^{23}}=0,5\ \text{mol}\\ n_{O_2}=\dfrac{6\cdot 10^{23}}{6\cdot 10^{23}}=1\ \text{mol}\\ n_{N_2}=\dfrac{9\cdot 10^{23}}{6\cdot 10^{23}}=1,5\ \text{mol}\)

\(\Rightarrow V_{O_2}=\dfrac{1}{0,5+1+1,5}\cdot 100\%=33,33\%\\ \%m_{O_2}=\dfrac{1\cdot 32}{0,5\cdot 44+1\cdot 32+1,5\cdot 28}\cdot 100\%=33,33\%\)

c. \(\%V_{O_2}=\dfrac{4}{2+4+6}\cdot 100\%=33,33\%\)

\(\%m_{O_2}=\dfrac{4\cdot 32}{2\cdot 44+4\cdot 32+6\cdot 28}\cdot 100\%=33,33\%\)