The density of ethanol,C₂H₅OH,is 0.789 g/mL.How many milliliters of ethanol are needed to produce 30.0 g of CO₂ according to the following chemical equation? C₂H₅OH(l)+ 3 O₂(g)→ 2 CO₂(g)+ 3 H₂O(l)

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First, we need to find the amount of CO₂ produced from 30.0 g of C₂H₅OH using stoichiometry. 30.0 g C₂H₅OH x (1 mol C₂H₅OH/46.07 g C₂H₅OH) x (2 mol CO₂/1 mol C₂H₅OH) = 1.304 mol CO₂ Next, we can use the density of ethanol to find the volume needed to produce that amount of CO₂. 1.304 mol CO₂ x (3 mol O₂/2 mol CO₂) x (22.4 L/1 mol) = 46.54 L O₂ 46.54 L O₂ x (1 L/1000 mL) x (0.789 g/mL) = 0.0367 g C₂H₅OH Finally, we can use the mass of C₂H₅OH needed and the density of ethanol to find the volume of ethanol needed. 30.0 g C₂H₅OH x (1 mL/0.789 g) = 38.03 mL Therefore, the answer is B) 19.9 mL (rounding to two significant figures).

The Density of Ethanol,C2H5OH,is 0

The Density of Ethanol,C₂H₅OH,is 0