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

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

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

PT: \(4Na+O_2\underrightarrow{t^o}2Na_2O\)

Xét tỉ lệ: \(\dfrac{0,5}{4}>\dfrac{0,1}{1}\), ta được Na dư.

a, Theo PT: \(n_{Na\left(pư\right)}=4n_{O_2}=0,4\left(mol\right)\)

\(\Rightarrow n_{Na\left(dư\right)}=0,1\left(mol\right)\Rightarrow m_{Na\left(dư\right)}=0,1.23=2,3\left(g\right)\)

b, Theo PT: \(n_{Na_2O}=2n_{O_2}=0,2\left(mol\right)\)

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

Bạn tham khảo nhé!

Answer 2

Ta có: \(\left\{{}\begin{matrix}n_{Na}=\dfrac{11,5}{23}=0,5\left(mol\right)\\n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\end{matrix}\right.\)

\(pthh:4Na+O_2\overset{t^o}{--->}2Na_2O\)

a. Ta thấy: \(\dfrac{0,5}{4}>\dfrac{0,1}{1}\)

Vậy Na dư.

Theo pt: \(n_{Na_{pứ}}=4.n_{O_2}=4.0,1=0,4\left(mol\right)\)

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

b. Áp dụng ĐLBTKL, suy ra:

\(m_{Na_2O}=0,4.23+3,2=12,4\left(g\right)\)