Hướng dẫn giải

Áp dụng ĐLBTKL ta có công thức khối lượng:

$\begin{align}& {{\text{m}}_{tr\ddot{o}\hat{o}\grave{u}c}}\text{=}{{\text{m}}_{\text{sau}}} \\& \text{mCa}{{\text{C}}_{\text{2}}}\text{+m}{{\text{H}}_{\text{2}}}\text{O=mCa(OH}{{\text{)}}_{\text{2}}}\text{+m}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}} \\& \text{m}{{\text{H}}_{\text{2}}}\text{O=mCa(OH}{{\text{)}}_{\text{2}}}\text{+m}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}\text{-mCa}{{\text{C}}_{\text{2}}} \\\end{align}$

= 13 + 37 – 41 = 9 gam.

Mặt khác ${{\text{V}}_{{{\text{H}}_{\text{2}}}\text{O}}}$  = m : d = 1= 9 : 1 = 9 ml. Chọn đáp án A.