Balanced equation:
2 C8H18 + 25 O2 16 CO2 + 18 H2O
Moles of Ocatane = mass / molar mass = 18.3 / 114 = 0.161 moles
Moles of O2 mass / molar mass = 84 / 32 = 2.625 moles
From the equation, mole sof O2 needed = moles of octane * 25 / 2 = 0.161 * 25 / 2 = 2.012 moles
Since there is more oxygen than needed, octane will completely react meaning miniumum leftover mass = 0 g
Limiting reactants Liquid octane (CH,(CH) CH,) will react with gaseous oxygen (0) to produce gaseous carbon...
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Liquid octane e CH(CH),CH,) will react with gaseous oxygen (0) to produce gaseous carbon dioxide (CO) and gaseous water (1,0). Suppose 10.39 of octane is mixed with 61. 9 of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 3 significant digits. X ? Explanation Check 30 MacBook Air 20 $ 4 % 5 6 7 8 2 3 U T R 14
Liquid octane (CH(CH),CH) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (1,0). S e 8.00 octane is mixed with 45.4 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 significant digits.
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