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4.35 L of Butane (C4H10) is combusted with 26.50 L of Oxygen gas. What volume of each product is formed by this reaction? What volume of excess reactant remains?
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...
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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: 2𝐶4𝐻10 + 13𝑂2 → 8𝐶𝑂2 + 10𝐻2𝑂
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= 4.615×44.01×0.59=119.83gMolar mass of H2O=18.01528Mole ratio H2O:O2 =10:13Mole ratio H2O: C4H10=10:25.77×18.01528×0.59=61.33g
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Based on the reaction below. How many grams of water would be form from 0.33 moles of C4H10 reacting with an unlimited amount of oxygen gas? (3pts) C4H10 + 6O2 --> 4CO2 + 5H2O
Mole ratio C4H10:H2O= 1:5= 0.33×5= 1.65molesMolar mass of H2O= 18.02= 18.02×1.65=29.733g
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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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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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Calculate the mass (in grams) of H2O produced from the reaction of 56.20 g C4H10 with 85.30 g O2.
C4H10 (g) + O2 (g) → CO2 (g) + H2O (l)
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
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Which liquid has the highest vapor pressure at 25°C?
Butane, C4H10
Glycerol, C3H5(OH)3
Octane, C8H18
Propanol, C3H7OH
The highest vapor pressure at 25 C has butane C4H10: 2.05 atm
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What volume, in liters ,is occupied by 8.15*10^21 molecules of butane (C4H10)?
under normal conditions it occupies a volume of 0.303 l
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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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Butane (C4H10) is burned with 35% excess air. Find the composition of the flue gas if combustion is complete.
2C4H10 + 13O2 = 8CO2 + 10H2Ofor burned 1 mol of butane, we need 6.5 mol of O2. If we have 35% excess than 1 mol of butane we waste...
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