Question

Giúp mình câu c

Answer 1

a) 

nếu như tính khối lượng MOL thì Mg=24(g) luôn

b)

\(n_{Al}=\dfrac{m}{M}=\dfrac{540}{27}=20\left(mol\right)\\ N_{Al}=20\cdot6\cdot10^{23}=1,2\cdot10^{25}\left(nguyentuAl\right)\)

c)

\(n_{CO_2}=\dfrac{1,5\cdot10^{23}}{6\cdot10^{23}}=0,25\left(mol\right)\\ m_{CO_2}=n\cdot M=0,25\cdot44=11\left(g\right)\)