Đáp án:
\(\begin{array}{l}
1.\\
a,{m_{N{a_2}O(dư)}} = 6,2g\\
b,{m_{NaCl}} = 11,7g\\
2.\\
a,{m_{HCl(dư)}} = 6,4g\\
b,{m_{MgC{l_2}}} = 1,1875g\\
c,\\
{m_{{H_2}}} = 0,025g\\
{V_{{H_2}}} = 0,28l\\
3.\\
{m_{Ca}} = 10g\\
{m_C} = 3g\\
{m_O} = 12g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1.\\
N{a_2}O + 2HCl \to 2NaCl + {H_2}O\\
{n_{N{a_2}O}} = 0,2mol\\
{n_{HCl}} = 0,2mol\\
a,\\
\to \dfrac{{{n_{N{a_2}O}}}}{1} > \dfrac{{{n_{HCl}}}}{2} \to {n_{N{a_2}O}}dư\\
\to {n_{N{a_2}O}} = \dfrac{1}{2}{n_{HCl}} = 0,1mol\\
\to {n_{N{a_2}O(dư)}} = 0,1mol\\
\to {m_{N{a_2}O(dư)}} = 6,2g\\
b,\\
{n_{NaCl}} = {n_{HCl}} = 0,2mol\\
\to {m_{NaCl}} = 11,7g\\
2.\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{m_{Mg}} = \dfrac{{1,806 \times {{10}^{23}}}}{{6,22 \times {{10}^{23}}}} = 0,3g\\
{n_{Mg}} = 0,0125mol\\
{n_{HCl}} = 0,2mol\\
\to {n_{HCl}} > {n_{Mg}} \to {n_{HCl}}dư\\
\to {n_{HCl}} = 2{n_{Mg}} = 0,025mol\\
\to {n_{HCl(du)}} = 0,175mol\\
a,\\
{m_{HCl(du)}} = 6,4g\\
b,\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,0125mol\\
\to {m_{MgC{l_2}}} = 1,1875g\\
c,\\
{n_{{H_2}}} = {n_{Mg}} = 0,0125mol\\
\to {m_{{H_2}}} = 0,025g\\
\to {V_{{H_2}}} = 0,28l\\
3.\\
{n_{CaC{O_3}}} = \dfrac{{25}}{{100}} = 0,25mol\\
\to {n_{Ca}} = {n_C} = {n_{CaC{O_3}}} = 0,25mol\\
\to {n_O} = 3{n_{CaC{O_3}}} = 0,75mol\\
\to {m_{Ca}} = 10g\\
\to {m_C} = 3g\\
\to {m_O} = 12g
\end{array}\)
