Đáp án:
\(\text C\%_{HCl\ \text{dư}}=7,684\%; \text C\%_{MgCl_2}=8,33\%\)
Giải thích các bước giải:
\(n_{MgO}=\dfrac{8}{40}=0,2\ \rm{mol}\)
\(m_{\text{dd HCl}}=200\cdot 1,1=220\ \rm{gam}\)
\(\to m_{\text{HCl}}=220\cdot 14,6\%=32,12\ \rm{gam}\)
\(\to n_{\rm{HCl}}=\dfrac{32,12}{36,5}=0,88\ \rm{mol}\)
\(MgO + 2HCl\to MgCl_2+H_2\)
\(\dfrac{0,2}1<\dfrac{0,88}2\to\)Sau phản ứng MgO hết
\(n_{\rm{HCl\ phản \ ứng}}=2\cdot n_{MgO}=0,4\ \rm{mol}\to n_{HCl\ \rm{dư}}=0,88-0,4=0,48\ \rm{mol}\to m_{HCl\ \rm{dư}}=36,5\cdot 0,48=17,52\ (gam)\)
\(n_{MgCl_2}=n_{\text{MgO}}=0,2\ \text{mol}\to m_{MgCl_2}=95\cdot 0,2=19\ \rm{gam}\)
BTKL: \(m_{\text{dd trước phản ứng}} =m_{\text{dd sau phản ứng}}\)
\(\to m_{\text{dd sau phản ứng}}=8+220=228\ \rm{gam}\)
\(\to \text C\%_{HCl\ dư}=\dfrac{17,52}{228}\cdot 100\%=7,684\%; \text C\%_{MgCl_2}=\dfrac{19}{228}\cdot 100\%=8,33\%\)