What volume of carbon dioxide measured at STP will be formed by the reaction of 1.47 mol of oxygen with 0.900 mol of ethyl alcohol,CH₃CH₂OH? A)40.3 mL B)22.0 L C)32.9 L D)49.4 L E)0.980 L

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The balanced chemical equation for the combustion of ethyl alcohol (C2H5OH) with oxygen (O2) is:$$ C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O $$From the equation, 1 mole of C2H5OH reacts with 3 moles of O2 to produce 2 moles of CO2. Given 0.900 mol of C2H5OH, the limiting reagent in this reaction is C2H5OH because it requires 2.7 moles of O2 (0.900 mol * 3), but only 1.47 mol of O2 is available. Therefore, the reaction is limited by the amount of O2.Since 3 moles of O2 produce 2 moles of CO2, 1.47 moles of O2 will produce:$$ \left( \frac{2}{3} \right) \times 1.47 = 0.98 \text{ moles of } CO_2 $$At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, 0.98 moles of CO2 will occupy:$$ 0.98 \times 22.4 \text{ L} = 21.952 \text{ L} $$Rounding to the nearest significant figure given in the question, the volume of CO2 formed is approximately 22.0 L.

What Volume of Carbon Dioxide Measured at STP Will Be