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Unit 11 Comprehensive Quiz - L Balance the following chemical equation: P_(4)+O_(2)arrow P_(2)O_(3) Use a 1 if no coefficient is needed. Coefficient for P_(4) square Coefficient for O_(2) square Coefficient for P_(2)O_(3) square

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Unit 11 Comprehensive Quiz - L Balance the following chemical equation: P_(4)+O_(2)arrow P_(2)O_(3) Use a 1 if no coefficient is needed. Coefficient for P_(4) square Coefficient for O_(2) square Coefficient for P_(2)O_(3) square

Unit 11 Comprehensive Quiz - L Balance the following chemical equation: P_(4)+O_(2)arrow P_(2)O_(3) Use a 1 if no coefficient is needed. Coefficient for P_(4) square Coefficient for O_(2) square Coefficient for P_(2)O_(3) square

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<br />Coefficient for $P_{4}$: 1<br />Coefficient for $O_{2}$: 3<br />Coefficient for $P_{2}O_{3}$: 2

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Balancing a chemical equation talks about making sure that the number of atoms of each element on the reactant side (left side of arrow) equals that on the product side (right side of arrow). In case of this question, we have the equation<br /><br />$P_{4}+O_{2}-P_{2}O_{3}$<br /><br />Our duty is to find out the coefficients for $P_{4}$, $O_{2}$, and $P_{2}O_{3}$ such that this equation gets balanced.<br /><br />1. Count the atoms of each element on both sides: We have the following:<br />P (Phosphorous): $4$ on left ($P_{4}$); and $2$ on right ($P_{2}O_{3}$);<br /><br />O (Oxygen): $2$ on left ($O_{2}$); and $3$ on right ($P_{2}O_{3}$);<br /><br /><br /><br />2. Start balance the equation from the Phosphorous since it shows up on left side as one compound ($P_{4}$). According to the number of P atoms, we need $2$ $P_{2}O_{3}$ on the right to balance things out.<br /><br /><br /><br />So we have to reassign the atoms: Now we have the equation:<br /><br />$P_{4}+O_{2}-2P_{2}O_{3}$;<br /><br />At this moment,<br />for Phosphorous $P$: $4$ on left; $4$ on right.<br /><br />for Oxygen $O$: $2$ on left:<br /> $6$ on the right.<br /><br /> <br /><br />3. To continue, just check the Oxygen balance staring from right side, we have $6$ Oxygen, which could just be obtained by simply putting $3$ in front of the Oxygen on left side ($O_{2}$). <br /><br /> Latest progression is :<br /> <br /> $P_{4}+3O_{2}-2P_{2}O_{3}$. <br /> <br /> This sums up the balancing. <br /><br />To collate this:<br /><br />Mean coefficient for Phosphorus is one: $1 \times P_{4} $; <br /><br />As determined in step 3 above; the coefficient for Oxygen is three: $3 \times O_{2} $;<br /><br />Finally, which we arrived at step 2 is the coefficient for $P_{2}O_{3}$ is two: $2 \times P_{2}O_{3}$ .<br /><br /><br />Transformation made by the determination phases makes the resultant equation to be:<br /><br />$P_{4}+ 3 O_{2} -2 P_{2}O_{3}$.
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