Ecuația reacției de neutralizare:
1 mol x
NaOH + HNO₃ -----> NaNO₃ + H₂O
1 1
Rezolvare:
[tex]m_s_1_{NaOH}=300g\\C_1=21\%\\\\C_1=\frac{m_d_1}{m_s_1}{\cdot}100\implies m_d_1=\frac{C_1{\cdot}m_s_1}{100}=\frac{21{\cdot}300}{100}=63\ g\ HNO_3\\\\M_{HNO_3}=1+14+(3{\cdot}16)=63\ g/mol\\\\n=\frac{m}{M}=\frac{63\ \not g}{63\ \frac{\not g}{mol}}=1\ mol\ HNO_3\\\\n_{NaNO_3}=x=1\ mol\\M_{NaNO_3}=23+14+(3{\cdot}16)=85\ g/mol[/tex]
[tex]n=\frac{m}{M}\implies m=n\ {\cdot}\ M=1 \not mol\ {\cdot}\ 85\ \frac{g}{\not mol}=85\ g\ NaNO_3\ (m_d_2)\\C_2=36.25\%\\\\C_2=\frac{m_d_2}{m_s_2}{\cdot}100\implies m_s_2=\frac{m_d_2{\cdot}100}{C}=\frac{85{\cdot}100}{36.25}=324.48\ g\ sol.\ NaNO_3\ adaugata[/tex]
