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burning of ethonal in air

C2H5OH burns in air:
C_2H_5OH(l)+O_2 --->CO_2(g)+H_2O(l)
balance the equation and determine the volume of air in liters at 35.0 C and 790 mmHg required to burn 227 g of ethanol. assume that air is 21.0 percent O_2 by volume.

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{In the chemical formulae below, please read the numbers as subscripts }
<br><br>I. Balancing the equation .....C2H5OH + O2 --> C02 + H20
<br><br>So this is easily done ... We have C2 on the Left Hand side(LHS) so we need atleast 2CO2 on the RHS to balance the Carbon. Also we have a total of (5+1)=6H on the LHS. To balance that we need atleast 3H20 on the RHS to balance the Hydrogen. So now, we just have to balance the Oxygen :on the RHS we have a total of (2*2+3*1 = 7) Oxygen. So to balance that we need 3O2 on the LHS. The balanced equation is :
<br><br>C2H5OH + 3 O2 --> 2 C02 + 3 H20
<br><br>II. Determining Volume
<br><br>As is usual we will work with ...

Solution Summary

It investigates the reaction of ethonal with O2, such as balancing the reaction and finding the amount of air needed.