balance the following equation and record the missing coefficients as your answer.\n__na + __o₂ →…

balance the following equation and record the missing coefficients as your answer.\n__na + __o₂ → __na₂o\nbalance the following equation and record the missing coefficients as your answer.\n__mg + __hcl → __mgcl₂ + __h₂\nbalance the following equation and record the missing coefficients as your answer.\n__nh₃ → __n₂ + __h₂
Answer
Explanation:
Step1: Balance the first equation
We need to balance the number of atoms on both sides of the equation $\text{Na}+\text{O}_2\rightarrow\text{Na}_2\text{O}$. There are 2 oxygen atoms on the left - hand side and 1 on the right - hand side. Multiply $\text{Na}_2\text{O}$ by 2 to get 2 oxygen atoms on the right. Then, to balance the sodium atoms, multiply $\text{Na}$ by 4. So the balanced equation is $4\text{Na}+\text{O}_2 = 2\text{Na}_2\text{O}$.
Step2: Balance the second equation
For the equation $\text{Mg}+\text{HCl}\rightarrow\text{MgCl}_2+\text{H}_2$, there are 2 chlorine atoms in $\text{MgCl}_2$. So we need 2 moles of $\text{HCl}$. Then the equation is balanced as $\text{Mg}+2\text{HCl}=\text{MgCl}_2+\text{H}_2$.
Step3: Balance the third equation
For the equation $\text{NH}_3\rightarrow\text{N}_2+\text{H}_2$, there are 2 nitrogen atoms in $\text{N}_2$. So multiply $\text{NH}_3$ by 2. Then there are 6 hydrogen atoms on the left - hand side, so multiply $\text{H}_2$ by 3. The balanced equation is $2\text{NH}_3=\text{N}_2 + 3\text{H}_2$.
Answer:
- 4, 1, 2
- 1, 2, 1, 1
- 2, 1, 3