Question

Answer 1

\(M_A=20,5.2=41g/mol\)

gọi \(\left\{{}\begin{matrix}x:n_{CO}\\y:n_{CO_2}\Rightarrow n_{SO_2}=y\end{matrix}\right.\)

Ta có:

\(M_A=\dfrac{m_A}{n_A}=\dfrac{28x+44y+64y}{x+y+y}=\dfrac{28x+108y}{x+2y}=41\)

<=> \(28x+108y=41\left(x+2y\right)\)

<=> \(28x+108y-41x-82y=0\)

<=> \(-13x=-26y\)

=> \(\dfrac{x}{y}=\dfrac{-26}{-13}=2\)

=> x = 2y

có: \(\%V_{CO}=\%n_{CO}=\dfrac{x.100}{x+y+y}=\dfrac{2y.100}{2y+y+y}=\dfrac{200y}{4y}=50\%\)

=> \(\%V_{CO_2}=\dfrac{y.100}{x+y+y}=\dfrac{y.100}{2y+y+Y}=\dfrac{100y}{4y}=25\%\)

=> \(\%V_{SO_2}=100-50-25=25\%\)