13 balance each of the following chemical equations.
a) ____ c(s) + ____ h₂(g) + ____ o₂(g) → ____ ch₃oh(g)
b) ____ c(s) + ____ cl₂(g) + ____ h₂(g) → ____ chcl₃(g)
c) ____ hgo(s) → ____ hg(l) + ____ o₂(g)
d) ____ li(s) + ____ o₂(g) → ____ li₂o(s)
e) ____ c₃h₈(g) + ____ o₂(g) → ____ co₂(g) + ____ h₂o(g)
f) ____ lioh(aq) + ____ co₂(g) → ____ li₂co₃(s) + ____ h₂o(l)

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Accepted Answer

# Explanation:
## Step1: Balance carbon atoms
For equation a), there is 1 C atom on the right (in $CH_3OH$) and 1 C atom on the left (in $C$). So the coefficient of $C$ is 1.
For equation b), there is 1 C atom on the right (in $CHCl_3$) and 1 C atom on the left (in $C$). So the coefficient of $C$ is 1.
For equation e), there are 3 C atoms on the left (in $C_3H_8$) and 1 C atom on the right (in $CO_2$). So the coefficient of $CO_2$ is 3.
## Step2: Balance hydrogen atoms
For equation a), there are 4 H atoms on the right (in $CH_3OH$) and 2 H atoms on the left (in $H_2$). So the coefficient of $H_2$ is 2.
For equation b), there is 1 H atom on the right (in $CHCl_3$) and 2 H atoms on the left (in $H_2$). So the coefficient of $H_2$ is 0.5. But we usually use whole - numbers, so multiply all coefficients by 2.
For equation c), there are 2 O atoms on the right (in $O_2$) and 1 O atom on the left (in $HgO$). So the coefficient of $HgO$ is 2.
For equation d), there are 2 O atoms on the left (in $O_2$) and 1 O atom on the right (in $Li_2O$). So the coefficient of $Li_2O$ is 2. Then for Li, there are 4 Li atoms on the right, so the coefficient of $Li$ is 4.
For equation e), there are 8 H atoms on the left (in $C_3H_8$) and 2 H atoms on the right (in $H_2O$). So the coefficient of $H_2O$ is 4.
For equation f), there are 2 Li atoms on the right (in $Li_2CO_3$) and 1 Li atom on the left (in $LiOH$). So the coefficient of $LiOH$ is 2.
## Step3: Balance oxygen atoms
For equation a), after balancing C and H, we have 1 O atom on the right (in $CH_3OH$) and 2 O atoms on the left (in $O_2$). So the coefficient of $O_2$ is 0.5. Multiply all coefficients by 2.
For equation c), after balancing Hg and O (from step 2), the equation is balanced.
For equation e), after balancing C and H, we have 3*2 + 4*1 = 10 O atoms on the right. There are 2 O atoms on the left (in $O_2$). So the coefficient of $O_2$ is 5.
For equation f), after balancing Li, we check O. On the left: 2 (from $LiOH$) + 2 (from $CO_2$) = 4. On the right: 3 (from $Li_2CO_3$) + 1 (from $H_2O$) = 4. And H: 2 (from $LiOH$) = 2 (from $H_2O$). The equation is balanced.

# Answer:
a) $2C_{(s)}+4H_{2(g)} + O_{2(g)}\to2CH_3OH_{(g)}$
b) $2C_{(s)}+6Cl_{2(g)}+2H_{2(g)}\to2CHCl_{3(g)}$
c) $2HgO_{(s)}\to2Hg_{(l)}+O_{2(g)}$
d) $4Li_{(s)}+O_{2(g)}\to2Li_2O_{(s)}$
e) $C_3H_{8(g)}+5O_{2(g)}\to3CO_{2(g)}+4H_{2O(g)}$
f) $2LiOH_{(aq)}+CO_{2(g)}\to Li_2CO_{3(s)}+H_{2O(l)}$

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QUESTION IMAGE

Question

Explanation:

Step1: Balance carbon atoms

Step2: Balance hydrogen atoms

Step3: Balance oxygen atoms

Answer: