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combustion of butane balanced equation how many moles of o2 are required to react completely with .35 mole of butane
7.83moles of Oxygen.
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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: 2πΆ4π»10 + 13π2 β 8πΆπ2 + 10π»2π
Molar mass of butane =58.1212/58.12=0.21molMolar mass of O2= 31.99812/31.998=0.38molMolar mass of CO2= 44.01Mole ratio CO2:O2= 8:13Mole ratio CO2: C4H10=8:2= 4.615Γ44.01Γ0.59=119.83gMolar mass of H2O=18.01528Mole ratio H2O:O2 =10:13Mole ratio H2O: C4H10=10:25.77Γ18.01528Γ0.59=61.33g
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1)Prepare n-butanyl bromide from n-butane
2)Prepare Ethyl bromide from n-butane
3)Prepare 1-Butanol from n-butane
4)Prepare 2-Butene from n-butane
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...
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Consider the compound A= 2 chloro 2 methy butane , B= 2 chloro butane, C= chloro propane. Which of the compound will readily undergo dehydrochlorination, what informs your answer? Expert's answer
The answer is compound A.If the mechanism of elimination is E1 then the reactivity of dehydrohalogenation is dependent on stability of carbocation. The more stable the carbocation the more readily...
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during combustion of butane at 225 degrees celsius a mixture of 5.00dm of butane and 75.0 dm of oxygen was ignite ,is there enough oxygen for the complete combustion?
5.00 dm3 x1 x2 2C4H10(g) + 13O2(g) rarr; 8CO2(g) + 10H2O(l) 2 13 8 Volume of O2 x1=5*13/2=32.5 Volume of CO2 x2=5*8/2=20 1.Yes ,we need 32.5 and we have 75
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Given 122 liters of butane gas,how many moles of butane are available for reaction?
n (C4H10) = V / 22.4 = 122 / 22.4 = 5.45 mol
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According to the following reaction, how many grams of bromine monochloride will be formed upon the complete reaction of 26.6 grams of chlorine gas with excess bromine?
The equation for this reaction is: Br2 + Cl2 β 2BrCl The number of moles of chlorine gas equals: Ξ½(Cl2) = m(Cl2)/Mr(Cl2) = 26.6 g / (2Γ35.5 g mol-1) =...
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Liquid bromine (Br2) has a density of 2.91g/mL. How many molecules of bromine are contained in a 8.50-mL sample of bromine?
m = VΓΟm(Br2) = 8.50 mL Γ 2.91 g/mL = 24.735 gM(Br2) = 159.808 g/moln = m/Mn(Br2) = 24.735 g / 159.808 (g/mol) = 0.154 molOne mole of Br2 molecules...
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A 79.9g sample of bromine contains approximately 1 mole of bromine atoms. How many bromine atoms are in a .80kg sample of bromine?
At first, kilogrammes should be turned into grams:m=80Γ1000=80000(g).Then, mass should be divided by molar mass:n=m/M=80000/79.9=1001.3(mol).(The data is not very clear. If the point is not mistake, and there are .80...
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Butane gas, C4H10, is combusted to produce carbon dioxide and water vapour. If 0.67 mol of butane are reacted, how many moles of water vapour are produced?
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
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