how many moles of h2o are produced when 64.0 grams c2h2 burns in oxygen?
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Answer Expert Verified Number of C₂H₂ moles reacted = 64.0 g / 26 g/mol = 2.46 mol.
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per mole of oxygen. From here, the answer is simple. Just multiply 2 by 31.9988 g because there are two moles of oxygen. So, the answer is 63.9976 g ..
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Therefore the molecular formula is twice the empirical formula: C2H6O2. The compound dioxane contains only carbon, hydrogen, and oxygen. When 0.956 g dioxane is burned, 1.91 g carbon dioxide and 0.782 g water are formed. In another experiment, it was determined that 6.04x10-3 mol dioxane weighs 0.532 g.
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Here is the balanced equation: 2C8H18(g)+25O2(g)→16CO2(g)+18H2O(g) number of moles of C8H18 reacted = 6.16×10−2 mol H2O produced = 0.554 mol oxygen is the limiting reactant.
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1 Answer. 4.73 mol O2 must be consumed to produce 68.1 g H2O in the given reaction.
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CH4+2O2======>CO2+2H20. So 1 mole of methane requires 2 moles of oxygen for complete combustion. So 16 g of Methane requires 64 g of oxygen for the complete combustion. So 8 g of Methane requires = 8x64/16=32 g of oxygen.
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Oxygen gas is O2 . O2 has a molar mass of 32g/mol . ∴ There are 1.25 moles of O2 in 40g of oxygen gas.
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Each mole of ethane undergo complete combustion with 3.5 moles of oxygen and produce 3 mole of water. Hence to produce 9 mole of water by combustion of 6 moles of ethane there's requirement of 10.5 mole of oxygen.
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Note: 27.66 g B₂H₆ is from the molar mass of diborane; it is necessary for us to use this to cancel the units. I provided an image for the cancellation of the necessary units. Hence, there will be 1.39 moles of water produced from 19.2 grams of B₂H₆.
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The equation says that for every 1O2 , there are 2H2O produced. Therefore, if you have 2.2O2 , you must have 2.2⋅2H2O , or 4.4 moles of H2O .
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The answer is 46.06844. We assume you are converting between grams C2H5OH and mole. You can view more details on each measurement unit: molecular weight of C2H5OH or mol The SI base unit for amount of substance is the mole. 1 grams C2H5OH is equal to 0.021706834440237 mole.
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The empirical formula of a certain hydrocarbon is CH2. When 0.125 moles of this hydrocarbon are completely combusted with excess oxygen, it is observed that 11.2 L of H2O gas is produced at STP. ... The ratio (0.500 / 0.125) = 4, and this is how many "H2" units are present in the hydrocarbon.
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Explanation: Clearly, four moles of water result from each mole of propane. And if 1.638⋅g of propane are combusted this represents a molar quantity of...
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According to the reaction: 2 mol of C2H6 produce 4 mol of CO2 5.70 mol of C2H6 ‒ x moles of CO2 So, number of moles of CO2 produced by combustion of 5.70 mol of ethane equals: Answer: n(CO2) = 11.4 mol.
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= 2,000 g of oxygen are needed to burn the 570 g (5 moles) of octane completely. The Brainliest Answer!
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How many moles of CO2 are produced when 0.400 mol of C6H12O6 reacts in this fashion? 0.400 mol C6H12O6 (2 mol CO2 / 1 mol C6H12O6) = 0.800 mol CO2 b.
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When you burn 6.25 moles of methane, you produce 6.25 moles of carbondioxide (275 grams of CO2 ) and 12.5 moles of water vapour (225 grams of H2O ).