Sửa đề: 132 → 13,2

Theo ĐL BTKL ta có:

\(m_{metan}+m_{oxi}=m_{cacbon}+m_{nước}\)

\(\Leftrightarrow4,8+m_{oxi}=13,2+10\)

\(\Leftrightarrow m_{oxi}=13,2+10-4,8=18,4\left(g\right)\)

\(PTHH:CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

Áp dụng ĐLBTKL ta có:

\(m_{CH_4}+m_{O_2}=m_{CO_2}+m_{H_2O}\)

\(\Rightarrow m_{O_2}=m_{CO_2}+m_{H_2O}-m_{CH_4}=132+10-4,8=137,2\left(g\right)\)

BTKL
m C + m O2 = m CO2

=> 24 + m O2 = 88

=> m O2 = 64 ( g )

\(n_{O_2}=\dfrac{8}{32}=0.25\left(mol\right)\)

\(CH_4+2O_2\underrightarrow{t^0}CO_2+2H_2O\)

\(0.125....0.25....0.125\)

\(m_{CH_4}=0.125\cdot16=2\left(g\right)\)

\(V_{CO_2}=0.125\cdot22.4=2.8\left(l\right)\)

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

\(............0.125.....0.125\)

\(m_{CaCO_3}=0.125\cdot100=12.5\left(g\right)\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
               0,2    0,4                  0,4 
\(V_{O_2}=0,4.22,4=8,96\left(l\right)\\ m_{H_2O}=0,4.18=7,2\left(g\right)\)

nCO2 = 4,48/22,4 = 0,2 (mol)

PTHH: CH4 + 2O2 -> (t°) CO2 + 2H2O

Mol: 0,2 <--- 0,4 <--- 0,2

VO2 = 0,4 . 22,4 = 8,96 (l)

Vkk = 8,96 . 5 = 44,8 (l)

mCH4 = 0,2 . 16 = 3,2 (g)

\(a,\text {Bảo toàn KL: }m_{C}+m_{O_2}=m_{CO_2}\\ \Rightarrow m_{CO_2}=m_{C}+m_{O_2}=16+6=22(g)\\ b,m_{C}=m_{CO_2}-m_{O_2}=44-32=12(g)\)

Câu 2 : 

Bảo toàn khối lượng : 

\(m_X=\dfrac{2.24}{22.4}\cdot44+3.6-\dfrac{4.48}{22.4}\cdot32=1.6\left(g\right)\)

Câu 1 :

\(CH_4+2O_2\underrightarrow{^{^{t^0}}}CO_2+2H_2O\)

Tỉ lệ số phân tử metan và oxi đã phản ứng = 1 : 2 

Bảo toàn khối lượng : 

\(m_{O_2}=4.4+1.8-1.6=4.6\left(g\right)\)

Câu 3 : 

Bảo toàn khối lượng 

\(m_{NaOH}=10.7+21.3-20=12\left(g\right)\)

\(n_{CO_2}=\dfrac{4.4}{44}=0.1\left(mol\right)\)

\(CH_4+2O_2\underrightarrow{^{^{t^0}}}CO_2+2H_2O\)

\(0.1.......0.2........0.1..........0.2\)

\(m_{CH_4}=0.1\cdot16=1.6\left(g\right)\)

\(V_{H_2O}=0.2\cdot22.4=4.48\left(l\right)\)

\(V_{kk}=5V_{O_2}=5\cdot0.2\cdot22.4=22.4\left(l\right)\)