Question

c_{10}h_{22}+o_{2}=>co_{2}+h_{2}o
al(oh)_{3}+hbr=>albr_{3}+h_{2}o
ch_{3}ch_{2}ch_{2}ch_{3}+o_{2}=>co_{2}+h_{2}o
c + o_{2}=>co_{2}
c_{3}h_{8}+o_{2}=>co_{2}+h_{2}o
li + alcl_{3}=>licl + al
c_{2}h_{6}+o_{2}=>co_{2}+h_{2}o
nh_{4}oh + h_{3}po_{4}=>(nh_{4})_{3}po_{4}+h_{2}o
rb + p=>rb_{3}p
ch_{4}+o_{2}=>co_{2}+h_{2}o
al(oh)_{3}+h_{2}so_{4}=>al_{2}(so_{4})_{3}+h_{2}o

Explanation:

Step1: Balance $C_{10}H_{22}+O_2

ightarrow CO_2 + H_2O$
First balance carbon and hydrogen. There are 10 carbons in $C_{10}H_{22}$, so we get $10CO_2$. There are 22 hydrogens, so we get $11H_2O$. Then balance oxygen: $2C_{10}H_{22}+31O_2 = 20CO_2+22H_2O$.

Step2: Balance $Al(OH)_3 + HBr

ightarrow AlBr_3+H_2O$
There are 3 bromines in $AlBr_3$, so we need 3 moles of $HBr$. Then balance oxygen and hydrogen: $Al(OH)_3 + 3HBr=AlBr_3 + 3H_2O$.

Step3: Balance $CH_3CH_2CH_2CH_3+O_2

ightarrow CO_2+H_2O$
There are 4 carbons in $CH_3CH_2CH_2CH_3$, so we get $4CO_2$. There are 10 hydrogens, so we get $5H_2O$. Then balance oxygen: $2CH_3CH_2CH_2CH_3+13O_2 = 8CO_2 + 10H_2O$.

Step4: Balance $C+O_2

ightarrow CO_2$
It is already balanced: $C + O_2=CO_2$.

Step5: Balance $C_3H_8+O_2

ightarrow CO_2+H_2O$
There are 3 carbons in $C_3H_8$, so we get $3CO_2$. There are 8 hydrogens, so we get $4H_2O$. Then balance oxygen: $C_3H_8+5O_2 = 3CO_2+4H_2O$.

Step6: Balance $Li+AlCl_3

ightarrow LiCl+Al$
There are 3 chlorines in $AlCl_3$, so we need 3 moles of $LiCl$. Then balance lithium: $3Li+AlCl_3 = 3LiCl+Al$.

Step7: Balance $C_2H_6+O_2

ightarrow CO_2+H_2O$
There are 2 carbons in $C_2H_6$, so we get $2CO_2$. There are 6 hydrogens, so we get $3H_2O$. Then balance oxygen: $2C_2H_6+7O_2 = 4CO_2+6H_2O$.

Step8: Balance $NH_4OH+H_3PO_4

ightarrow (NH_4)_3PO_4+H_2O$
There are 3 ammonium ions in $(NH_4)_3PO_4$, so we need 3 moles of $NH_4OH$. Then balance oxygen and hydrogen: $3NH_4OH + H_3PO_4=(NH_4)_3PO_4+3H_2O$.

Step9: Balance $Rb+P

ightarrow Rb_3P$
There are 3 rubidiums in $Rb_3P$, so we need 3 moles of $Rb$: $3Rb + P=Rb_3P$.

Step10: Balance $CH_4+O_2

ightarrow CO_2+H_2O$
There is 1 carbon in $CH_4$, so we get $1CO_2$. There are 4 hydrogens, so we get $2H_2O$. Then balance oxygen: $CH_4+2O_2 = CO_2+2H_2O$.

Step11: Balance $Al(OH)_3+H_2SO_4

ightarrow Al_2(SO_4)_3+H_2O$
There are 2 aluminums in $Al_2(SO_4)_3$, so we need 2 moles of $Al(OH)_3$. There are 3 sulfate groups in $Al_2(SO_4)_3$, so we need 3 moles of $H_2SO_4$. Then balance oxygen and hydrogen: $2Al(OH)_3+3H_2SO_4 = Al_2(SO_4)_3+6H_2O$.

Answer:

$2C_{10}H_{22}+31O_2 = 20CO_2+22H_2O$; $Al(OH)_3 + 3HBr=AlBr_3 + 3H_2O$; $2CH_3CH_2CH_2CH_3+13O_2 = 8CO_2 + 10H_2O$; $C + O_2=CO_2$; $C_3H_8+5O_2 = 3CO_2+4H_2O$; $3Li+AlCl_3 = 3LiCl+Al$; $2C_2H_6+7O_2 = 4CO_2+6H_2O$; $3NH_4OH + H_3PO_4=(NH_4)_3PO_4+3H_2O$; $3Rb + P=Rb_3P$; $CH_4+2O_2 = CO_2+2H_2O$; $2Al(OH)_3+3H_2SO_4 = Al_2(SO_4)_3+6H_2O$