reactions · reactions in solution (balancing equations)
introduction laboratory simulation
type the correct coefficient for each reactant and product.
1) balance the chemical equation for the reaction of aqueous sodium hydroxide and aqueous copper(ll) sulfate.
$ \square \mathrm { naoh } ( a q ) + \square \mathrm { cuso } _ { 4 } ( a q ) \
ightarrow \square \mathrm { cu } ( \mathrm { oh } ) _ { 2 } ( s ) + \square \mathrm { na } _ { 2 } \mathrm { so } _ { 4 } ( a q ) $
2) balance the chemical equation for the reaction of aqueous sodium chloride and aqueous lead(ll) nitrate.
$ \square \mathrm { nacl } ( a q ) + \square \mathrm { pb } ( \mathrm { no } _ { 3 } ) _ { 2 } ( a q ) \
ightarrow \square \mathrm { pbcl } _ { 2 } ( s ) + \square \mathrm { nano } _ { 3 } ( a q ) $
3) balance the chemical equation for the reaction of aqueous barium chloride and sulfuric acid.
$ \square \mathrm { bacl } _ { 2 } ( a q ) + \square \mathrm { h } _ { 2 } \mathrm { so } _ { 4 } ( a q ) \
ightarrow \square \mathrm { baso } _ { 4 } ( s ) + \square \mathrm { hcl } ( a q ) $

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Accepted Answer

# Explanation:
## Step1: Balance the first equation
For $ \text{NaOH}(aq)+\text{CuSO}_4(aq)\to\text{Cu(OH)}_2(s)+\text{Na}_2\text{SO}_4(aq) $
- Sodium ($Na$): On the left - hand side, there is $1$ $Na$ in $NaOH$, and on the right - hand side, there are $2$ $Na$ in $Na_2SO_4$.
- Hydroxide ($OH$): On the left - hand side, there is $1$ $OH$ in $NaOH$, and on the right - hand side, there are $2$ $OH$ in $Cu(OH)_2$.
- Copper ($Cu$) and sulfate ($SO_4$) are already balanced.
- If we put a coefficient of $2$ in front of $NaOH$, the equation becomes $2\text{NaOH}(aq)+\text{CuSO}_4(aq)\to\text{Cu(OH)}_2(s)+\text{Na}_2\text{SO}_4(aq)$

## Step2: Balance the second equation
For $ \text{NaCl}(aq)+\text{Pb(NO}_3)_2(aq)\to\text{PbCl}_2(s)+\text{NaNO}_3(aq) $
- Chloride ($Cl$): On the left - hand side, there is $1$ $Cl$ in $NaCl$, and on the right - hand side, there are $2$ $Cl$ in $PbCl_2$.
- Sodium ($Na$) and nitrate ($NO_3$): After balancing $Cl$ by putting a coefficient of $2$ in front of $NaCl$, we have $2$ $Na$ on the left. To balance $Na$ and $NO_3$ (since there are $2$ $NO_3$ in $Pb(NO_3)_2$), we put a coefficient of $2$ in front of $NaNO_3$. The equation becomes $2\text{NaCl}(aq)+\text{Pb(NO}_3)_2(aq)\to\text{PbCl}_2(s)+2\text{NaNO}_3(aq)$

## Step3: Balance the third equation
For $ \text{BaCl}_2(aq)+\text{H}_2\text{SO}_4(aq)\to\text{BaSO}_4(s)+\text{HCl}(aq) $
- Barium ($Ba$) and sulfate ($SO_4$) are already balanced.
- Chloride ($Cl$): On the left - hand side, there are $2$ $Cl$ in $BaCl_2$, and on the right - hand side, there is $1$ $Cl$ in $HCl$.
- Hydrogen ($H$): On the left - hand side, there are $2$ $H$ in $H_2SO_4$.
- If we put a coefficient of $2$ in front of $HCl$, the equation becomes $ \text{BaCl}_2(aq)+\text{H}_2\text{SO}_4(aq)\to\text{BaSO}_4(s)+2\text{HCl}(aq) $

# Answer:
1. $2\text{NaOH}(aq)+\text{CuSO}_4(aq)\to\text{Cu(OH)}_2(s)+\text{Na}_2\text{SO}_4(aq)$
2. $2\text{NaCl}(aq)+\text{Pb(NO}_3)_2(aq)\to\text{PbCl}_2(s)+2\text{NaNO}_3(aq)$
3. $ \text{BaCl}_2(aq)+\text{H}_2\text{SO}_4(aq)\to\text{BaSO}_4(s)+2\text{HCl}(aq) $

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QUESTION IMAGE

Question

Explanation:

Step1: Balance the first equation

Step2: Balance the second equation

Step3: Balance the third equation

Answer: