Đáp án:
a) 6,4 g
b) 2,24 l
Giải thích các bước giải:
\(\begin{array}{l}
2Ca + {O_2} \to 2CaO\\
2Mg + {O_2} \to 2MgO\\
mCa = 0,625m\,g\\
mMg = 0,375m\,g\\
nCa = \dfrac{{0,625m}}{{40}} = \dfrac{m}{{64}}\,mol\\
= > nCaO = \dfrac{m}{{64}}\,mol\\
nMg = \dfrac{{0,375m}}{{24}} = \dfrac{m}{{64}}\,mol\\
= > nMgO = \dfrac{m}{{64}}\,mol\\
\dfrac{m}{{64}} \times 56 + \dfrac{m}{{64}} \times 40 = 9,6\\
= > m = 6,4g\\
b)\\
nCa = \dfrac{{6,4}}{{64}} = 0,1\,mol\\
nMg = \dfrac{{6,4}}{{64}} = 0,1\,mol\\
= > n{O_2} = 0,05 + 0,05 = 0,1\,mol\\
V{O_2} = 0,1 \times 22,4 = 2,24l
\end{array}\)