6. Balance the reaction below. If it goes to completion, what total volume will it occupy at STP? Will the volume have increased or decreased?By how much? (Assume ideality.) C_(3)H_(6)O(l)+4O_(2)(g)arrow 3CO_(2)(g)+3H_(2)O(g)

To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides of the equation. In this case, the reaction is already balanced as written:<br /><br />$C_{3}H_{6}O(l)+4O_{2}(g)\rightarrow 3CO_{2}(g)+3H_{2}O(g)$<br /><br />Now, let's calculate the total volume occupied at STP (Standard Temperature and Pressure). At STP, 1 mole of an ideal gas occupies 22.4 liters.<br /><br />First, we need to determine the number of moles of each reactant and product:<br /><br />1. $C_{3}H_{6}O(l)$: 1 mole<br />2. $O_{2}(g)$: 4 moles<br />3. $CO_{2}(g)$: 3 moles<br />4. $H_{2}O(g)$: 3 moles<br /><br />Now, we calculate the total volume occupied by each substance at STP:<br /><br />1. $C_{3}H_{6}O(l)$: Since it is a liquid, it occupies negligible volume.<br />2. $O_{2}(g)$: 4 moles * 22.4 L/mol = 89.6 L<br />3. $CO_{2}(g)$: 3 moles * 22.4 L/mol = 67.2 L<br />4. $H_{2}O(g)$: 3 moles * 22.4 L/mol = 67.2 L<br /><br />The total volume occupied by the gases at STP is:<br /><br />89.6 L + 67.2 L + 67.2 L = 224 L<br /><br />Since the reaction goes to completion, the total volume occupied by the products will be the same as the total volume occupied by the reactants. Therefore, the volume has increased by 224 L.