Question

Đố𝓽 𝓬𝓱á𝔂 11,5𝓰 𝓝𝓪 𝓽𝓻𝓸𝓷𝓰 𝓫ì𝓷𝓱 𝓬𝓱ứ𝓪 3,2𝓰 𝓴𝓱í 𝓞2.𝓢𝓪𝓾 𝓹𝓱ả𝓷 ứ𝓷𝓰 𝓬𝓱ấ𝓽 𝓷à𝓸 𝓬ò𝓷 𝓭ư

𝓪.𝓣í𝓷𝓱 𝓶 𝓬𝓱ấ𝓽 𝓭ư

𝒷.𝒯í𝓃𝒽 𝓂 𝓈ả𝓃 𝓅𝒽ẩ𝓂 𝓈𝓃𝒽 𝓇𝒶

Answer 1

4Na+O₂-to>2Na₂O

           0,1-----0,2 mol

n _Na=\(\dfrac{11,5}{23}\)=0,5 mol

n _O2=\(\dfrac{3,2}{32}\)=0,1 mol

=>Na dư , dư 0,025 mol

=>m _Nadư=0,025.23=0,575 g

=> m _{sản phẩm}  = 0,2.62=12,4g

Answer 2

a.

\(n_{Na}=\dfrac{11,5}{23}=0,5\left(mol\right)\)

\(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\)

PTHH : 4Na + O₂ -> 2Na₂O

              0,4     0,1         0,2 ( mol ) 

Xét tỉ lệ :  \(\dfrac{0,5}{4}>\dfrac{0,1}{1}\)=> Na dư , O₂ đủ

\(m_{Na\left(dư\right)}=\left(0,5-0,4\right).23=2,3\left(g\right)\)

b) \(m_{Na_2O}=0,2.62=12,4\left(g\right)\)