The answer is 9.94 ml.
Given,
The density of ethanol, C2H5OH = 0.789 g/mL
[tex]n (CO_{2} ) = \frac{m}{M} = \frac{15G}{44 g/mol} } = 0.341 mol;[/tex]
[tex]n ( C_{2} H_{2} OH) = \frac{n (CO_{2}) }{2} = \frac{0.341}{2} = 0.1705 mol;[/tex]
[tex]m (C_{2} H_{2} OH) = 0.1705 mol * 46 g/ mol[/tex] = 7.843 g
[tex]V (C_{2} H_{2} OH ) = \frac{7.843}{0.789} = 9.94 ml.[/tex]
Therefore, the answer is 9.94 ml
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The complete question is -
If the density of ethanol, C2H5OH, is 0.789 g/mL. How many milliliters of ethanol are needed to produce 15.0 g of CO2 according to the following chemical equation?
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)