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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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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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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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How many moles of barium metal react to produce 0.100mol of barium nitride? Hpw many moles of nitrogren gas react?
Ba(s) + N2(g) -> Ba3N2(s)
x y 0.100 3Ba(s) + 2N2(g) ------gt; Ba3N2(s) 3 2 1 x=3*0.100/1=0.300 mol y=2*0.100/2-0.200 mol
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How many moles of oxygen are needed to burn 9.3 moles of butane in this combustion reaction?
The chemical equation for the burning reaction is 2C4H10 + 13O2 = 8CO2 + 10H2O According to the chemical equation n(O2) = n(C4H10) : 2 x 13 = 9.3 :...
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Butane is used as a fuel in camping gas stoves.100 cm3 of butane is burned in excess oxygen .calculate the volume of carbon dioxide produced at r.t.p
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...
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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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In the reaction 4Al(s)+3O2(g)-> Al2O3(s)
A. How many moles of oxygen react with 5.5 miles of Al?
B. How many moles of Al2O3 can be produced from 5.5 moles of Al?
A. 4,125B. 1,375
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The density of butane, C4H10 , is 2.49×10−3 g/mL . Calculate the number of moles of butane in a 35.9 mL sample.
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
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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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