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The symbol equation for combustion of a hydrocarbon is shown below. What number will go before the oxygen reactant when this equation is balanced? Enter your answer as a number C_(3)H_(8)+ldots O_(2)arrow ldots CO_(2)+ldots H_(2)O

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The symbol equation for combustion of a hydrocarbon is shown below. What number will go before the oxygen reactant when this equation is balanced? Enter your answer as a number C_(3)H_(8)+ldots O_(2)arrow ldots CO_(2)+ldots H_(2)O

The symbol equation for combustion of a hydrocarbon is shown below. What number will go before the oxygen reactant when this equation is balanced? Enter your answer as a number C_(3)H_(8)+ldots O_(2)arrow ldots CO_(2)+ldots H_(2)O

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YaraMaster · Tutor for 5 years

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<p> 5</p>

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<p> This problem requires knowledge of how to balance equations in chemistry, specifically considering the number of atoms on both sides of a chemical equation. One side represents the reactants, whereas the other symbolises the products. <br /><br />Chemical compounds involved in this equation are: Propane or \(\mathrm{C}_{3} \mathrm{H}_{8}\), Oxygen, \(\mathrm{O}_{2}\), Carbon Dioxide \(\mathrm{CO}_{2}\) and water, \(\mathrm{H}_{2} \mathrm{O}\). <br /><br />Looking at the equation, the way to balance such an equation is to ensure an equal number of each type of atom on each side. According to the above unbalanced equation, we can identify that, not considering the values we need to find, there are three carbons, 8 hydrogens, and 2 oxygens at the left side and at the right side, there exist carbons and oxygens in \(\mathrm{CO}_{2}\) and two Hydrogen atoms and one oxygen atom in \(\mathrm{H}_{2} \mathrm{O}\).<br /><br />We know that combustion reactions require sufficient amount of Oxygen to completely burn a substance.<br /><br />If we get started with balancing carbons, then for there being 3 carbons on the left side, there should also be 3 on the right. That is why, the equation turns to<br />\[ <br />\mathrm{C}_{3} \mathrm{H}_{8}+...O_2 \rightarrow 3...CO_2 + ...H_2O <br />\]<br /><br />Next, for having Carbon equation satisfied and looking to Hydrogen we can observe that, to imbalance Hydrogen atoms, we ought to have 8 Hydrogen at each side. Therefore, multiplication with 4 obtain this balance.<br />\[ <br />\mathrm{C}_{3} \mathrm{H}_{8}+...O_2 \rightarrow 3CO_2 +4H_2O <br />\]<br /><br />After this crucial step, we look at balancing the number of oxygen atoms. After balancing Carbon and Hydrogen atoms, the total number of Oxygen becomes: from \(3CO_2\) that means = \(3*2=6\), and from \(4H_2O = 4*1=4\) so the total Oxygen on the right side = 6 + 4 = 10.<br /><br />Here, it is important to achieve such that an even number of total Oxygen exists. Since oxygen gas in its molecular form mostly stands as \(\mathrm{O}_{2}\), on the right side there ought to me a coefficient makes the total number of oxygen ten units.<br /><br />The last step then allows us to complete from where we simultaneously satisfy the balance between atoms. Once this lesion is followed consistently we obtain the balanced chemical reaction:<br />\[ <br />\mathrm{C}_{3} \mathrm{H}_{8}+5...O_2 \rightarrow 3CO_2 +4H_2O. <br />\]<br />Therefore, The number that will go with the Oxygen Reactor (Oxygen Gas) is 5.<br /></p>
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