Đáp án:
\(\begin{array}{l}
a)\\
C{\% _{C{H_3}COOH}} = 24\% \\
b)\\
{m_{{C_4}{H_{10}}}} = 34,8g
\end{array}\)
Giải thích các bước giải:
500g dung dịch $NaHCO_3$ 20% nha em
\(\begin{array}{l}
a)\\
C{H_3}COOH + NaHC{O_3} \to C{H_3}COONa + C{O_2} + {H_2}O\\
{m_{NaHC{O_3}}} = \dfrac{{500 \times 20}}{{100}} = 100g\\
{n_{NaHC{O_3}}} = \dfrac{{100}}{{84}}1,2mol\\
{n_{C{H_3}COOH}} = {n_{NaHC{O_3}}} = 1,2mol\\
{m_{C{H_3}COOH}} = 1,2 \times 60 = 72g\\
C{\% _{C{H_3}COOH}} = \dfrac{{72}}{{300}} \times 100\% = 24\% \\
b)\\
2{C_4}{H_{10}} + 5{O_2} \to 4C{H_3}COOH + 2{H_2}O\\
{n_{{C_4}{H_{10}}}} = \dfrac{{{n_{C{H_3}COOH}}}}{2} = 0,6mol\\
{m_{{C_4}{H_{10}}}} = 0,6 \times 58 = 34,8g
\end{array}\)
