How do you balance $N a (s) + N a N O_{3} (a q) \to N a_{2} O (a q) + N_{2} (g)$ $Na(s) + NaNO_3(aq) -> Na_2O(aq) + N_2(g)$ ?

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How do you balance $N a (s) + N a N O_{3} (a q) \to N a_{2} O (a q) + N_{2} (g)$ $Na(s) + NaNO_3(aq) -> Na_2O(aq) + N_2(g)$ ?

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Answer 1 · 2016-03-05T03:47:55

$10 Na (s) + 2 {NaNO}_{3} (a q) \to 6 {Na}_{2} O (a q) + N_{2} (g)$ $10"Na"(s) + 2"NaNO"_3(aq) -> 6"Na"_2"O"(aq) + "N"_2(g)$

Explanation:

After balancing the equation, you should see that the number of $Na$ $"Na"$ , $O$ $"O"$ and $N$ $"N"$ should be the same on both sides of the equation. This is the original equation.

$Na + {NaNO}_{3} \to {Na}_{2} O + N_{2}$ $"Na" + "NaNO"_3 -> "Na"_2"O" + "N"_2$

When balancing the equations, I omit the state symbols. I will add them back in the final balanced equation.

The trick to balancing the equation is to fix the compound with the most elements present to a ratio of $1$ $1$ . In this case, it is ${NaNO}_{3}$ $"NaNO"_3$ , as it contains all 3 elements.

Next, we start of by balancing the elements that have the least occurrence. $Na$ $"Na"$ appeared in 3 times, while $N$ $"N"$ and $O$ $"O"$ appeared only 2 times, so we start with $N$ $"N"$ first.

$Balancing N$ $color(green)"Balancing N"$

There is 1 $N$ $"N"$ on the left-hand side (LHS) and 2 $N$ $"N"$ on the right-hand side (RHS). To balance $N$ $"N"$ , we only need half the amount of $N_{2}$ $"N"_2$ on the RHS.

$Na + {NaNO}_{3} \to {Na}_{2} O + \frac{1}{2} N_{2}$ $"Na" + "NaNO"_3 -> "Na"_2"O" + 1/2 "N"_2$

$Balancing O$ $color(green)"Balancing O"$

There are 3 $O$ $"O"$ on the LHS and 1 $O$ $"O"$ on the RHS. To balance $O$ $"O"$ , we need 3 times the amount of ${Na}_{2} O$ $"Na"_2"O"$ on the RHS.

$Na + {NaNO}_{3} \to 3 {Na}_{2} O + \frac{1}{2} N_{2}$ $"Na" + "NaNO"_3 -> 3"Na"_2"O" + 1/2 "N"_2$

$Balancing Na$ $color(green)"Balancing Na"$

There are 2 $Na$ $"Na"$ on the LHS and 6 $Na$ $"Na"$ on the RHS. Both the number of ${NaNO}_{3}$ $"NaNO"_3$ and ${Na}_{2} O$ $"Na"_2"O"$ are fixed already, which means that we can only change the number of $Na$ $"Na"$ . To balance $Na$ $"Na"$ , we need 5 more $Na$ $"Na"$ on the LHS.

$5 Na + {NaNO}_{3} \to 3 {Na}_{2} O + \frac{1}{2} N_{2}$ $5"Na" + "NaNO"_3 -> 3"Na"_2"O" + 1/2 "N"_2$

The balancing is now complete. However, some people do not like fractions as coefficients. So to get rid of the $\frac{1}{2}$ $1/2$ , multiply everything by $2$ $2$ . It becomes

$10 Na + 2 {NaNO}_{3} \to 6 {Na}_{2} O + N_{2}$ $10"Na" + 2"NaNO"_3 -> 6"Na"_2"O" + "N"_2$

Remember to put back the state symbols!

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How do you balance $N a (s) + N a N O_{3} (a q) \to N a_{2} O (a q) + N_{2} (g)$ $Na(s) + NaNO_3(aq) -> Na_2O(aq) + N_2(g)$ ?

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How do you balance Na(s)+NaNO3(aq)→Na2O(aq)+N2(g)Na(s) + NaNO_3(aq) -> Na_2O(aq) + N_2(g)?

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How do you balance $N a (s) + N a N O_{3} (a q) \to N a_{2} O (a q) + N_{2} (g)$ $Na(s) + NaNO_3(aq) -> Na_2O(aq) + N_2(g)$ ?