Question

Đốt cháy hoàn toàn 6g Magie (Mg) trog oxi thu dc Magie Oxit (MgO). tính khối lượng của Magie oxit (MgO) thu dc?

Answer 1

2Mg + O₂ \(\underrightarrow{to}\) 2MgO

\(n_{Mg}=\frac{6}{24}=0,25\left(mol\right)\)

Theo pT: \(n_{MgO}=n_{Mg}=0,25\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,25\times40=10\left(g\right)\)

Answer 2

2Mg + O2 -t-> 2MgO

n_Mg = 0,25(mol)=n_MgO

=> m_MgO = 0,25.40=10(g)

Answer 3

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

2 1 2

0,25 0,125 0,25

\(n_{Mg}=\frac{6}{24}=0,25\left(mol\right)\)

→\(n_{MgO}=n_{Mg}=0,25\left(mol\right)\)

➝\(m_{MgO}=0,25.\left(24+16\right)=10\left(g\right)\)

Answer 4

nMg = 6/24 = 0.25 mol

Mg + 1/2O2 -to-> MgO

0.25____________0.25

mMgO = 0.25*40= 10g

Answer 5

\(n_{Mg}=0,25\left(mol\right)\\ PTHH:Mg+\frac{1}{2}O_2\underrightarrow{t^o}MgO\)

(mol) 1 1

(mol) 0,25 0,25

\(m_{MgO}=n.M=0,25.40=10\left(g\right)\)

Answer 6

PTHH: Mg+ O---(t^o)---> MgO

............6............................10...........(g)

Vậy thu được 10 g MgO.

Answer 7

https://i.imgur.com/wl5zDhq.jpg

Answer 8

2Mg + O₂ --> 2MgO

n_Mg = 6/24 = 0,25 mol

n_MgO = n_Mg = 0,25 mol

m_MgO = 0,25.40 = 10 g