12) ___ na + ___ h₂o → ___ naoh + ___h₂
13) ___ ag₂o → ___ ag + ___o₂
14) ___ s₈ + ___o₂ → ___ so₃
15) ___ co₂ + ___ h₂o → ___ c₆h₁₂o₆ + ___o₂
16) ___ k + ___ mgbr → ___ kbr + ___ mg
17) ___ hcl + ___ caco₃ → ___ cacl₂ + ___h₂o + ___ co₂
18) ___ hno₃ + ___ nahco₃ → ___ nano₃ + ___ h₂o + ___ co₂
19) ___ h₂o + ___ o₂ → ___ h₂o₂
20) ___ nabr + ___ caf₂ → ___ naf + ___ cabr₂
21) ___ h₂so₄ + ___ nano₂ → ___ hno₂ + ___ na₂so₄

Question

12)  ___ na + ___ h₂o → ___ naoh + ___h₂
13)  ___ ag₂o → ___ ag + ___o₂
14)  ___ s₈ + ___o₂ → ___ so₃
15)  ___ co₂ + ___ h₂o → ___ c₆h₁₂o₆ + ___o₂
16)  ___ k + ___ mgbr → ___ kbr + ___ mg
17)  ___ hcl + ___ caco₃ → ___ cacl₂ + ___h₂o + ___ co₂
18)  ___ hno₃ + ___ nahco₃ → ___ nano₃ + ___ h₂o + ___ co₂
19)  ___ h₂o + ___ o₂ → ___ h₂o₂
20)  ___ nabr + ___ caf₂ → ___ naf + ___ cabr₂
21)  ___ h₂so₄ + ___ nano₂ → ___ hno₂ + ___ na₂so₄

Accepted Answer

# Explanation:
## Step1: Balance equation 12
For $Na + H_2O
ightarrow NaOH + H_2$, balance $H$ first. There are $2$ $H$ in $H_2O$ and $3$ $H$ in products ($1$ in $NaOH$ and $2$ in $H_2$). Multiply $H_2O$ by $2$, $NaOH$ by $2$, then $Na$ by $2$. So $2Na + 2H_2O = 2NaOH + H_2$.
## Step2: Balance equation 13
For $Ag_2O
ightarrow Ag + O_2$, balance $O$. Multiply $Ag_2O$ by $2$, $O_2$ by $1$, then $Ag$ by $4$. So $2Ag_2O = 4Ag + O_2$.
## Step3: Balance equation 14
For $S_8 + O_2
ightarrow SO_3$, balance $S$ first. Multiply $SO_3$ by $8$, then $O_2$ by $12$. So $S_8 + 12O_2 = 8SO_3$.
## Step4: Balance equation 15
For $CO_2 + H_2O
ightarrow C_6H_{12}O_6 + O_2$, balance $C$ first. Multiply $CO_2$ by $6$. Then balance $H$, $H_2O$ by $6$. Then balance $O$, $O_2$ by $6$. So $6CO_2 + 6H_2O = C_6H_{12}O_6 + 6O_2$.
## Step5: Balance equation 16
For $K + MgBr
ightarrow KBr + Mg$, balance $Br$ first. Multiply $K$ by $1$, $KBr$ by $1$. So $K + MgBr = KBr + Mg$.
## Step6: Balance equation 17
For $HCl + CaCO_3
ightarrow CaCl_2 + H_2O + CO_2$, balance $Cl$ first. Multiply $HCl$ by $2$. So $2HCl + CaCO_3 = CaCl_2 + H_2O + CO_2$.
## Step7: Balance equation 18
For $HNO_3 + NaHCO_3
ightarrow NaNO_3 + H_2O + CO_2$, count atoms. $1$ $H$, $1$ $N$, $3$ $O$ in $HNO_3$; $1$ $Na$, $1$ $H$, $1$ $C$, $3$ $O$ in $NaHCO_3$. Products: $1$ $Na$, $1$ $N$, $3$ $O$ in $NaNO_3$; $2$ $H$, $1$ $O$ in $H_2O$; $1$ $C$, $2$ $O$ in $CO_2$. So $1HNO_3 + 1NaHCO_3 = 1NaNO_3 + 1H_2O + 1CO_2$.
## Step8: Balance equation 19
For $H_2O + O_2
ightarrow H_2O_2$, balance $O$. Multiply $H_2O$ by $2$, $H_2O_2$ by $2$. So $2H_2O + O_2 = 2H_2O_2$.
## Step9: Balance equation 20
For $NaBr + CaF_2
ightarrow NaF + CaBr_2$, balance $Br$ and $F$. Multiply $NaBr$ by $2$, $NaF$ by $2$. So $2NaBr + CaF_2 = 2NaF + CaBr_2$.
## Step10: Balance equation 21
For $H_2SO_4 + NaNO_3
ightarrow HNO_3 + Na_2SO_4$, balance $Na$. Multiply $NaNO_3$ by $2$, $HNO_3$ by $2$. So $H_2SO_4 + 2NaNO_3 = 2HNO_3 + Na_2SO_4$.

# Answer:
12. $2Na + 2H_2O = 2NaOH + H_2$
13. $2Ag_2O = 4Ag + O_2$
14. $S_8 + 12O_2 = 8SO_3$
15. $6CO_2 + 6H_2O = C_6H_{12}O_6 + 6O_2$
16. $K + MgBr = KBr + Mg$
17. $2HCl + CaCO_3 = CaCl_2 + H_2O + CO_2$
18. $HNO_3 + NaHCO_3 = NaNO_3 + H_2O + CO_2$
19. $2H_2O + O_2 = 2H_2O_2$
20. $2NaBr + CaF_2 = 2NaF + CaBr_2$
21. $H_2SO_4 + 2NaNO_3 = 2HNO_3 + Na_2SO_4$

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QUESTION IMAGE

Question

Explanation:

Step1: Balance equation 12

Step2: Balance equation 13

Step3: Balance equation 14

Step4: Balance equation 15

Step5: Balance equation 16

Step6: Balance equation 17

Step7: Balance equation 18

Step8: Balance equation 19

Step9: Balance equation 20

Step10: Balance equation 21

Answer: