Question

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Mn ơi giúp mình giải với

Answer 1

[Nguyễn Ngọc Huy Toàn]

Áp dụng ĐLBTKL, ta có:

\(m_{O_2}=36,1-23,3=12,8g\)

\(\rightarrow V_{O_2}=\dfrac{12,8}{32}.22,4=8,96l\)

Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\\n_{Zn}=y\end{matrix}\right.\)

\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)

  x       1/2 x               x          ( mol )

\(2Zn+O_2\rightarrow\left(t^o\right)2ZnO\)

 y        1/2 y              y          ( mol )

Ta có:

\(\left\{{}\begin{matrix}24x+65y=23,3\\\dfrac{1}{2}x+\dfrac{1}{2}y=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,7\\y=0,1\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,7.24=16,8g\\m_{Zn}=0,1.65=6,5g\end{matrix}\right.\)

Answer 2

[Buddy]

3)

2Mg+O2-to>2MgO

x-------1\2x----------x

2Zn+O2-to>2ZnO

y-------1\2y------------y

ta có \(\left\{{}\begin{matrix}24x+65y=23,3\\40x+81y=36,1\end{matrix}\right.\)

=>x=0,7  mol, y=0,1 mol

=>VO2=0,4.22,4=8,96l

=>m Mg=24.0,7=16,8g

=>m Zn=65.01,=6,5g