The chemical equation shows the combustion of propane (C3H8). cap c sub 3 cap h sub 8 plus 5 cap o sub 2 rightarrow 3 CO sub 2 plus 4 cap h sub 2 cap o The molar mass of oxygen gas open paren cap o sub 2 close paren is 32 point 0 0 g over mol The molar mass of cap c sub 3 cap h sub 8 is 44 point 1 g over mol What mass of cap o sub 2 , in grams, is required to completely react with 0 point 0 2 5 g cap c sub 3 cap h sub 8 ?
The balanced equation shows that 1 mole of propane reacts with 5 moles of oxygen. The propane amount is 0.025 g ÷ 44.1 g/mol = 0.000567 mol, so it requires 5 × 0.000567 = 0.00284 mol O₂. Converting to mass gives 0.00284 mol × 32.00 g/mol = 0.091 g O₂, which is choice C.
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