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

Gộp cả phần a và b

Ta có: \(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,25mol\\n_{MgO}=0,5mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{O_2}=0,25\cdot22,4=5,6\left(l\right)\\m_{MgO}=0,5\cdot40=20\left(g\right)\end{matrix}\right.\)

a, Mg+O2​to-->2MgO

b,

nMg​=0,5(mol)

nO2=0,25 mol;nMgO=0,5mol

→;nMgO​=0,5(mol)

→VO2​=5,6(l)

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

b)

\(n_{Mg} = \dfrac{2,4}{24} = 0,1(mol)\)

Theo PTHH : 

\(n_{O_2} = \dfrac{1}{2}n_{Mg} = 0,05(mol)\\ \Rightarrow V_{O_2} = 0,05.22,4 = 1,12(lít)\)

c)

\(n_{MgO} = n_{Mg} = 0,1(mol)\\ \Rightarrow m_{MgO} = 0,1.40 = 4(gam)\)

d)

\(V_{không\ khí} = 5V_{O_2} = 1,12.5 = 5,6(lít)\)

a, 2Mg + O₂ \(\underrightarrow{t^o}\) 2MgO

b, \(n_{Mg}=\dfrac{4,8}{24}=0,2mol\)

\(n_{O_2}=\dfrac{0,2}{2}=0,1mol\)

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

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

c, Cách 1:

\(Theo.ĐLBTKL,ta.có:\\ m_{Mg}+m_{O_2}=m_{MgO}\)

\(\Rightarrow m_{MgO}=4,8+3,2=8g\)

Cách 2:

\(n_{MgO}=\dfrac{0,2.2}{2}=0,2mol\)

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

BTKL: \(m_{Mg}+m_{O_2}=m_{MgO}\)

Do đó: \(m_{O_2}=m_{MgO}-m_{Mg}=30-18=12(g)\)

Theo ĐLBT KL, có: m_Mg + m_O2 = m_MgO

⇒ m_O2 = m_MgO - m_Mg = 30 - 18 = 12 (g)

Bạn tham khảo nhé!

n_Mg = 9,6/24 = 0,4 (mol)

2Mg + O2 ---t^o---> 2MgO

0,4____0,2_________0,4

V_O2(đktc) = 0,2.22,4 = 4,48(l)

m_MgO = 0,4.40 = 16(g)

nMg = 1.2/24 = 0.05 (mol) 

2Mg + O2 -to-> 2MgO 

0.05__0.025____0.05 

mMgO = 0.05*40 = 2 (g) 

VO2 = 0.025*22.4 = 0.56(l)

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

b/ Ta có: \(n_{Mg}=\dfrac{1.2}{24}=0.05\left(mol\right)\)

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

2                   2

0.05              x

\(=>x=0.05=n_{MgO}\)

=> \(m_{MgO}=0.05\cdot\left(24+16\right)=2\left(g\right)\)

\(a) 2Mg + O_2 \xrightarrow{t^o} 2MgO\\ b)\\ n_{MgO} = n_{Mg} = \dfrac{1,2}{24} = 0,05(mol)\\ \Rightarrow m_{MgO} = 0,05.40 = 2(gam)\\ c)\\ n_{O_2} = \dfrac{1}{2}n_{Mg} = 0,025(mol)\\ \Rightarrow V_{O_2} = 0,025.22,4 = 0,56(lít)\)

a) mMg + mO2 = mMgO.

b) mO2= mMgO – mMg = 15 - 9 = 6(g).

a) m M g + m O 2 = m M g O 

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

Bảo toàn khối lượng: \(m_{Mg}+m_{O_2}=m_{MgO}\)

\(\Rightarrow m_{O_2}=m_{MgO}-m_{Mg}=4-2,4=1,6\left(g\right)\)