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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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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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Gas barbecues burn propane using oxygen from the air. If 5.00 L of propane is burned, predict the volume of oxygen, at the same temperature and pressure, required for complete combustion.
Solution:Propane (C3H8).The balanced chemical equation:C3H8 + 5O2 = 3CO2 + 4H2OAccording to the equation: n(C3H8) = n(O2)/5n(gas) = V(gas) / VmHence,V(C3H8)/Vm = V(O2)/5VmV(C3H8) = V(O2)/5V(O2) = 5 × V(C3H8) =...
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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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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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Determine the amount of theoretical oxygen and theoretical air needed to burn
a.) 60 kg of butane (C4H10)
b.) 20 kg of propane (C5H12)
c.) 80 kg of octane (C8H18)
A. 2C4H10 + 13O2 ----> 8CO2 + 10H2OMolar mass of C4H10 = (12x4)+10= 58gmol-1molar mass of O2= 16x2=32gmol-12(58)g of C4H10 required 13(32)g of OxygenTherefore 60Kg of C4H10 will require 416x60/116...
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-Hexane (C6H14) is being used as a fuel. The Moles of Oxygen required for complete combustion of 1 Mole of the fuel is: Select one: a. 9.5 moles b. 10.25 moles c. 10 moles d. None of the above
C6H14+9.5O2=6CO2 +7H2OAnswer a - 9.5 mole
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