\(n_{MgO}=\frac{10}{40}=0,25\left(mol\right)\)
\(m_{HCl}=\frac{115.26\%}{100\%}=29,9\left(g\right)\)
\(n_{HCl}=\frac{29,9}{36,5}\approx0,82\left(mol\right)\)
\(PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\)
Ban đầu: \(0,25\)_____\(0,82\)
Phản ứng: \(0,25\)____\(0,5\)____\(0,25\)______\(0,25\) \(\left(mol\right)\)
Dư:______________\(0,32\)
Lập tỉ lệ: \(\frac{0,25}{1}< \frac{0,82}{2}\left(0,25< 0,41\right)\)
\(\Rightarrow MgO\) hết \(HCl\) dư
Các chất sau phả ứng là \(HCl\left(dư\right)\) và \(MgCl_2\)
\(m_{H_2O}=0,25.18=4,5\left(g\right)\)
\(m_{ddsaupư}=m_{MgO}+m_{ddHCl}+m_{H_2O}=10+115+4,5=129,5\left(g\right)\)
\(m_{HCl\left(dư\right)}=0,32.36,5=11,68\left(g\right)\)
\(C\%_{HCl\left(dư\right)}=\frac{11,68}{129,5}.100\%=9\%\)
\(m_{MgCl_2}=0,25.95=23,75\left(g\right)\)
\(C\%_{MgCl_2}=\frac{23,75}{129,5}.100\%=18,34\%\)
