Question

instructions - balance the chemical equations by changing the coefficients of different chemicals.

1. $\boldsymbol{1}$f$_2$ + $\boldsymbol{2}$nacl $

ightarrow$ $\boldsymbol{2}$naf + $\boldsymbol{1}$cl$_2$
2 f x2
2 x na x2
2 x cl 2

1. _na$_3$po$_4$ + _al(no$_3$)$_3$ $

ightarrow$ ___nano$_3$ + ___alpo$_4$
na
po
al
no$_3$

1. _ch$_4$ + _o$_2$ $

ightarrow$ ___co$_2$ + ___h$_2$o

1. _s$_8$ + _o$_2$ $

ightarrow$ ___so$_3$

Explanation:

Step1: Balance equation 2 (Na first)

Count Na: Left has 3 (in $\text{Na}_3\text{PO}_4$), so set coefficient of $\text{NaNO}_3$ to 3.
$\underline{1}\text{Na}_3\text{PO}_4 + \underline{1}\text{Al(NO}_3\text{)}_3
ightarrow \underline{3}\text{NaNO}_3 + \underline{1}\text{AlPO}_4$
Verify all atoms: Na(3=3), P(1=1), Al(1=1), $\text{NO}_3$(3=3)

Step2: Balance equation 3 (C first)

C is balanced (1=1). Balance H: Left has 4 (in $\text{CH}_4$), so set $\text{H}_2\text{O}$ coefficient to 2.
$\underline{1}\text{CH}_4 + \underline{2}\text{O}_2
ightarrow \underline{1}\text{CO}_2 + \underline{2}\text{H}_2\text{O}$
Verify O: Right has $2 + 2=4$, left has $2\times2=4$

Step3: Balance equation 4 (S first)

Left has 8 S (in $\text{S}_8$), so set $\text{SO}_3$ coefficient to 8.
$\underline{1}\text{S}_8 + \underline{12}\text{O}_2
ightarrow \underline{8}\text{SO}_3$
Verify O: Right has $8\times3=24$, left has $12\times2=24$

Answer:

1. $\boldsymbol{\underline{1}\text{F}_2 + \underline{2}\text{NaCl}

ightarrow \underline{2}\text{NaF} + \underline{1}\text{Cl}_2}$ (already balanced)

1. $\boldsymbol{\underline{1}\text{Na}_3\text{PO}_4 + \underline{1}\text{Al(NO}_3\text{)}_3

ightarrow \underline{3}\text{NaNO}_3 + \underline{1}\text{AlPO}_4}$

1. $\boldsymbol{\underline{1}\text{CH}_4 + \underline{2}\text{O}_2

ightarrow \underline{1}\text{CO}_2 + \underline{2}\text{H}_2\text{O}}$

1. $\boldsymbol{\underline{1}\text{S}_8 + \underline{12}\text{O}_2

ightarrow \underline{8}\text{SO}_3}$