Question

[Soái Ca Mới]

Hòa tan 4,4g hỗn hợp bột X gồm Mg và MgO bằng một lượng vừa đủ dd HCl 14,6% thu được 2,24 lít khí ở đktc

a-Tính % theo khối lượng từng chất trong hỗn hợp X

b- Tính % dd sau phản ứng

Answer 1

[Kiêm Hùng]

Đặt \(n_{Mg}=x;n_{MgO}=y\)

\(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\)

(mol) 1 2 1 1

(mol) x 2x x x

\(PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\)

(mol) 1 2 1 1

(mol) y 2y y y

\(hpt:\left\{{}\begin{matrix}24x+40y=4,4\\22,4x=2,24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_{Mg}=0,1\left(mol\right)\rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\rightarrow\%m_{Mg}=\frac{2,4}{4,4}.100\%=54,54\left(\%\right)\\n_{MgO}=0,05\left(mol\right)\rightarrow m_{MgO}=0,05.40=2\left(g\right)\rightarrow\%m_{MgO}=100\%-54,54\%=45,46\left(\%\right)\end{matrix}\right.\)

\(m_{ddHCl}=\frac{36,5.\left(2.0,1+2.0,05\right).100\%}{14,6\%}=75\left(g\right)\)

\(m_{H_2}=x.2=0,1.2=0,2\left(g\right)\)

\(C\%_{ddspu}=\frac{\left(0,1+0,05\right).95}{4,4+75-0,2}.100\%=18\left(\%\right)\)

Answer 2

[Phùng Hà Châu]

Mg + 2HCl → MgCl₂ + H₂ (1)

MgO + 2HCl → MgCl₂ + H₂O (2)

\(n_{H_2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)

a) Theo Pt1: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,1\times24=2,4\left(g\right)\)

\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)

\(\%m_{Mg}=\frac{2,4}{4,4}\times100\%=54,55\%\)

\(\%m_{MgO}=\frac{2}{4,4}\times100\%=45,45\%\)

b) \(m_{H_2}=0,1\times2=0,2\left(g\right)\)

Theo pT1: \(n_{HCl}=2n_{Mg}=2\times0,1=0,2\left(mol\right)\)

\(n_{MgO}=\frac{2}{40}=0,05\left(mol\right)\)

Theo Pt2: \(n_{HCl}=2n_{MgO}=2\times0,05=0,1\left(mol\right)\)

\(\Rightarrow\Sigma n_{HCl}=0,2+0,1=0,3\left(mol\right)\)

\(\Rightarrow\Sigma m_{HCl}=0,3\times36,5=10,95\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\frac{10,95}{14,6\%}=75\left(g\right)\)

Ta có: \(m_{dd}saupứ=4,4+75-0,2=79,2\left(g\right)\)

Theo pt1: \(n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\)

Theo Pt2: \(n_{MgCl_2}=n_{MgO}=0,05\left(mol\right)\)

\(\Rightarrow\Sigma n_{MgCl_2}=0,1+0,05=0,15\left(mol\right)\)

\(\Rightarrow\Sigma m_{MgCl_2}=0,15\times95=14,25\left(g\right)\)

\(\Rightarrow C\%_{MgCl_2}=\frac{14,25}{79,2}\times100\%=17,99\%\)