Consider the reaction below. What is the limiting reactant when 52.0 g of ethene (C2H4, molar mass = 28.0 g/mol) and 128 g of oxygen (molar mass = 32.0 g/mol) are reacted? C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(g) • CO2 • H2O • C2H4 • O2
Added by Tammy P.
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0 g and the molar mass of 28.0 g/mol. Moles of C2H4 = 52.0 g / 28.0 g/mol = 1.857 moles ** Show more…
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