If 12 grams of butane (C4H10), a fuel in cigarette lighter, reacts with 12 grams of O2, how many grams of the products will be formed? Balanced Equation: 2𝐶4𝐻10 + 13𝑂2 → 8𝐶𝑂2 + 10𝐻2𝑂

7.83moles of Oxygen.

2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g)Since the question specified that the combustion reaction of butane (C4H10), represented by the chemical equation in the question, took place at S.T.P. (standard temperature and pressure conditions), then we...

Mole ratio C4H10:H2O= 1:5= 0.33×5= 1.65molesMolar mass of H2O= 18.02= 18.02×1.65=29.733g

1) CH3-CH2-CH2-CH3 + Cl2 ---gt; CH3-CH2-CH2-CH2-Cl + HCl CH3-CH2-CH2-CH2-Cl + KOH (dry) --gt; CH3-CH2-CH2=CH2 CH3-CH2-CH2=CH2 + Br2/H2O2 --gt; CH3-CH2-CH2-CH2-Br CH3-CH2-CH2-CH2-Br + KOH (aq) --gt; CH3-CH2-CH2-CH2-OH + KBr CH3-CH2-CH2-CH2-OH + NaBr/H2SO4...

2 C4H10 + 13 O2 = 8 CO2 + 10 H2O n = V/22.4 n (C4H10) = 0.1/22.4 = 0.0047 mol n (CO2) = 4 · n (C4H10) = 4...

C4H10(g)  + 6,5O2(g) → 4CO2(g) + 5H2O(l)n(C4H10)=m(C4H10)/M(C4H10)=56.20/58=0.969moln(O2)=m(O2)/M(O2)=85.30/32=2.666moln(C4H10)/1 and n(O2)/6.5 0.969/1 and 2.666/6.5 0.969 and 0.43The mass of water is calculated by the deficiency, by n(O2)n(O2)/n(H2O)=6.5/52.666*5=6.5*n(H2O)n(H2O)=2.05molV(H2O)=n(H2O)*Vm=2.05*22.4=45.92l=45920mlρ(H2O)=1g/mlm(H2O)=ρ*V(H2O)=1*45920=45920gAnswer - 45920g

The highest vapor pressure at 25 C has butane C4H10: 2.05 atm

 2.49×10−3 g/mL × 35.9 mL = 0.089391 g Butane molar mass 58.124 g·mol−1 0.089391 / 58.124 g·mol−1 = 0.001538 mol

2C4 H10 + 13 O2 = 8 CO2 +10 H2On (C4 H10 )= 0,67 mole n (H2O)=10*0.67/2=3,35 moleanswer: 3,35 mole

Cigarettes smoke contains carbon-bashed solid particles suspended in the gas produce on combustion of cigarettes.These carbon-bashed solid particles make the smoke visible.In the colloid if cigarettes smokeDispersed phase : Solid...