Đáp án:
b) 60g
c) 11,2 l
d) 82g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2C{H_3}COOH + 2Na \to 2C{H_3}COONa + {H_2}\\
b)\\
nNa = \dfrac{m}{M} = \dfrac{{23}}{{23}} = 1\,mol\\
nC{H_3}COOH = nNa = 1\,mol\\
mC{H_3}COOH = n \times M = 1 \times 60 = 60g\\
c)\\
n{H_2} = \dfrac{1}{2} = 0,5\,mol\\
V{H_2} = n \times 22,4 = 0,5 \times 22,4 = 11,2l\\
d)\\
nC{H_3}COONa = nNa = 1\,mol\\
mC{H_3}COONa = n \times M = 1 \times 82 = 82g
\end{array}\)
