Question

Đố𝓽 𝓬𝓱á𝔂 4,8𝓰 𝓜𝓰 𝓽𝓻𝓸𝓷𝓰 𝓫ì𝓷𝓱 𝓬𝓱ứ𝓪 4,48 𝓵𝓲𝓽 𝓞2 (đ𝓴𝓽𝓬)

𝓪.𝓣í𝓷𝓱 𝓥 𝓬𝓱ấ𝓽 𝓭ư 𝓼𝓪𝓾 𝓹ư?

𝓫.𝓣í𝓷𝓱 𝓶 𝓼ả𝓷 𝓹𝓱ẩ𝓶 𝓼𝓲𝓷𝓱 𝓻𝓪 ?

Answer 1

Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PT: \(2Mg+O_2\underrightarrow{t^o}2MgO\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,2}{1}\), ta được O₂ dư.

a, Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{Mg}=0,1\left(mol\right)\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,1\left(mol\right)\Rightarrow V_{O_2\left(dư\right)}=0,1.22,4=2,24\left(l\right)\)

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

\(\Rightarrow m_{MgO}=0,2.40=8\left(g\right)\)

Bạn tham khảo nhé!

Answer 2

a. 

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH : 2Mg + O₂ -> 2MgO

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

Xét tỉ lệ : \(\dfrac{0,2}{2}< \dfrac{0,2}{1}\) => Mg đủ , O₂ dư

\(n_{O_2\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

\(m_{O_2\left(dư\right)}=0,1.32=3,2\left(g\right)\)

\(V_{O_2\left(dư\right)}=0,1.22,4=2,24\left(l\right)\)

b. 

\(m_{MgO}=0,2.40=8\left(g\right)\)

 

Answer 3

Answer 4

\(a,n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ PTHH:2Mg+O_2\rightarrow\left(t^o\right)2MgO\\ Vì:\dfrac{0,2}{2}< \dfrac{0,2}{1}\Rightarrow Mghết,O_2dư\\ n_{O_2\left(p.ứ\right)}=\dfrac{1}{2}.n_{Mg}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ \Rightarrow n_{O_2\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\\ \Rightarrow V_{O_2\left(dư,đktc\right)}=0,1.22,4=2,24\left(l\right)\\ b,n_{MgO}=n_{Mg}=0,2\left(mol\right)\\ \Rightarrow m_{MgO}=0,2.40=8\left(g\right)\)