\(C_2H_2+\frac{5}{2}O_2\underrightarrow{t^0}2CO_2+H_2O\)

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

a, \(n_{CH_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH:

CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

0,5--->1------------->0,5

Ca(OH)₂ + CO₂ ---> CaCO₃ + H₂O

                   0,5----->0,5

b, \(V_{O_2}=1.22,4=22,4\left(l\right)\)

c, \(m_{CaCO_3}=0,5.100=50\left(g\right)\)

\(1.C_2H_5OH+3O_2-^{t^o}\rightarrow2CO_2+3H_2O\left(1\right)\\ 2.n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\\ TheoPT\left(1\right):n_{O_2}=3n_{C_2H_5OH}=0,6\left(mol\right)\\ \Rightarrow V_{kk}=\dfrac{0,6.22,4}{20\%}=67,2\left(l\right)\\ 3.CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\left(2\right)\\TheoPT\left(1\right):n_{CO_2}=2n_{C_2H_5OH}=0,4\left(mol\right)\\TheoPT\left(2\right) n_{CO_2}=n_{CaCO_3}=0,4\left(mol\right)\\ \Rightarrow x=0,4.100=40\left(g\right)\\4.Độrượu=\dfrac{V_{rượunguyenchat}}{V_{ddrượu}}.100\\ V_{ruowujnguyenchat\:}=\dfrac{m}{D}=\dfrac{9,2}{0,8}=12\left(ml\right)\\ \Rightarrow V_{ddrượu20^o}=\dfrac{12}{20}.100=60\left(ml\right)\)

\(n_{CaCO_3}=\dfrac{25}{100}=0.25\left(mol\right)\)

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

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

\(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0.25}{0.1}=2.5\left(M\right)\)

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

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

\(V_{CH_4}=0.25\cdot22.4=5.6\left(l\right)\)

a) nCaCO3=0,25(mol)

CH4 + 2 O2 -to-> CO2 + 2 H2O

0,25<------------------0,25(mol)

CO2 + Ca(OH)2 -> CaCO3 + H2O

0,25<------0,25-----------0,25(mol)

b) CMddCa(OH)2= 0,25/0,1= 2,5(M)

b) V(CH4,đktc)=0,25.22,4=5,6(l)

a) Hiện tượng hóa học

b) Hiện tượng vật lí

c) Hiện tượng vật lí

d) Hiện tượng hóa học

a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

b, \(n_{CH_4}=\dfrac{3}{16}=0,1875\left(mol\right)\)

Theo PT: \(n_{CO_2}=n_{CH_4}=0,1875\left(mol\right)\Rightarrow m_{CO_2}=0,1875.44=8,25\left(g\right)\)

c, \(n_{CaCO_3}=n_{CO_2}=0,1875\left(mol\right)\Rightarrow m_{CaCO_3}=0,1875.100=18,75\left(g\right)\)

CH₄ + 2H₂O  CO₂ + 4H₂

CH₄ + 2O₂ (kk)  CO₂ + 2H₂O nên còn lại N₂

N₂ + 3H₂ ⇔ 2NH₃

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

Ta có: \(n_{CH_4}=\dfrac{3,2}{16}=0,2\left(mol\right)\)

Theo PT: \(n_{O_2}=2n_{CH_4}=0,4\left(mol\right)\)

\(\Rightarrow V_{O_2}0,4.22,4=8,96\left(l\right)\)

b, Theo PT: \(n_{CO_2}=n_{CH_4}=0,2\left(mol\right)\)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

Theo PT: \(n_{CaCO_3}=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{\downarrow}=m_{CaCO_3}=0,2.100=20\left(g\right)\)

Bạn tham khảo nhé!

a) khí hidro + khí oxi -> nước b) vôi sống + nước -> vôi tôi c) kali clorat -> kali clorua + khí oxi d) đá vôi + khí oxi-> vôi sống + khí cacbonic e) lưu huỳnh + khí oxi -> khí sunfuro f) kẽm + axit clohydric -> kẽm clorua + khí hidro. g) bari clorua + axit sunfuric -> bari sunfat + axit clohidric