# Balance the following reaction: CuCl2 + NaNO3  Cu(NO3)2 + NaCl If 15 grams of copper (II) chloride react with 20 grams of sodium nitrate, how much sodium chloride can be formed? What is the name of the limiting reagent? How much of the excess reagent is left over in this reaction?

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86
Dec 3, 2014

For this question we have to start with balanced a chemical equation.

$C u C {l}_{2}$ + $2 N a N {O}_{3}$ $\rightarrow$ $C u {\left(N {O}_{3}\right)}_{2}$ + $2 N a C l$

Formula:
Number of moles = $\text{mass in grams"/"molar mass}$

Molar mass of $C u C {l}_{2}$ = 134.45g/mol
Molar mass of $N a N {O}_{3}$ = 85.00g/mol
Molar mass of $C u {\left(N {O}_{3}\right)}_{2}$ = 187.57g/mol
Molar mass of $N a C l$ = 58.44g/mol

To determine the limiting reagent and the amount of excess reagent left over in the reaction:

Amount of $C u C {l}_{2}$ present(in moles) = $\text{15"/"134.45}$ = 0.11156 mol $\approx$ 0.11 mol (2 significant figures)
Amount of $N a N {O}_{3}$ present(in moles) = $\text{20"/"85.00}$ = 0.23529 mol $\approx$ 0.24 mol (2 significant figures)

From the above chemical equation,
1 mol of 1 mol of $C u C {l}_{2}$ will react with 2 mol of $N a N {O}_{3}$
Therefore,
0.11156 mol of $C u C {l}_{2}$ will react with 0.22312 mol of $N a N {O}_{3}$

Number of mole of excess $N a N {O}_{3}$ = 0.23529 - 0.22312 = 0.01217
Amount of excess $N a N {O}_{3}$(in grams) = 0.01217 $\cdot$ 85.00 = 1.03445g $\approx$ 1.034g (4 significant figures)

Hence, since sodium nitrate is present in excess, copper (II) chloride is the limiting reagent.

To find the mass of sodium chloride produced from this reaction:

From the above chemical equation.
1 mol of $C u C {l}_{2}$ reacts to produce 2 mol of $N a C l$
Therefore,
0.11156 mol of $C u C {l}_{2}$ will react to produce 0.22312 mol of $N a C l$

Mass of $N a C l$ produced = 0.22312 $\cdot$ 58.44 = 13.0391328g $\approx$ 13.04g (4 significant figures)

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Meave60 Share
Dec 15, 2014

The reaction "CuCl"_2("aq") + "2NaNO"_3("aq") $\rightarrow$ ${\text{Cu(NO"_3)}}_{2} \left(a q\right)$ + $\text{2NaCl(aq)}$ does not occur because all of the ions are in solution. It should be written as "CuCl"_2("aq") + "2NaNO"_3("aq") $\rightarrow$ $\text{no reaction}$ or $\text{N.R.}$. There is no limiting reagent or reagent in excess.

However, the reaction "CuCl"_2("aq") + $\text{2NaOH(aq)}$ $\rightarrow$ "Cu(OH)"_2("s") + $\text{2NaCl(aq)}$ does occur because one of the products, "Cu(OH)"_2("s") is a precipitate.

The limiting reagent and reagent in excess can be determined for this reaction.

Just as an exercise, I will use 15g of copper (II) chloride and 20g of sodium hydroxide.

In order to solve this problem, we will need to use mole ratios of the reactants and products. The coefficients in front of each compound gives us the number of moles of each compound. We also need the molar masses for copper (II) chloride and sodium hydroxide. We also need the molar mass for one of the products. I'm going to choose sodium chloride.

Known/Given:
molar mass of ${\text{CuCl}}_{2}$ = $\text{134.45 g/mol}$ (Wikipedia)
molar mass of $\text{NaOH}$ = $\text{39.997 g/mol}$ (Wikipedia)
molar mass of $\text{NaCl}$ = $\text{58.44 g/mo}$l (Wikipedia)

Equation:

"CuCl"_2("aq") + $\text{2NaOH(aq)}$ $\rightarrow$ "Cu(OH)"_2("s") + $\text{2NaCl(aq)}$

Solution:

First, determine the limiting reagent by determining the mass of the product $\text{NaCl}$ possible for each reactant. Use the molar mass of each reactant, the mole ratio between each reactant and $\text{NaCl}$ from the balanced equation, and the molar mass of $\text{NaCl}$.

${\text{15g CuCl}}_{2}$x$\text{1mol CuCl2"/"134.45g CuCl2}$x$\text{2mol NaCl"/"1mol CuCl2}$x$\text{58.44g NaCl"/"1 mol NaCl}$=$\text{13g NaCl}$

$\text{20g NaOH}$x$\text{1mol NaOH"/"39.997g NaOH}$x$\text{2mol NaCl"/"2mol NaOH}$x$\text{58.44g NaCl"/"1mol NaCl}$=$\text{30g NaCl}$

The limiting reagent is ${\text{CuCl}}_{2}$.

The reagent in excess is $\text{NaOH}$.

Second, determine the amount of excess reagent that did react with ${\text{CuCl}}_{2}$. Use the molar mass of ${\text{CuCl}}_{2}$, the mole ratio between $\text{NaOH}$ and ${\text{CuCl}}_{2}$ from the balanced equation, and the molar mass of $\text{NaOH}$.

${\text{15g CuCl}}_{2}$x$\text{1mol CuCl2"/"134.45g CuCl2}$x$\text{2mol NaOH"/"1mol CuCl2}$x$\text{39.997g NaOH"/"1mol NaOH}$=$\text{8.9g NaOH}$

Third, determine how much $\text{NaOH}$ was in excess. Subtract the amount that did react from the amount that you started with.

Excess $\text{NaOH}$ = $\text{20g NaOH - 8.9g NaOH = 11g NaOH}$ (rounded to a whole number because of $\text{20g}$)

Conclusion:
The maximum amount of $\text{NaCl}$ that can be produced by this reaction is $\text{13 g}$.

The limiting reagent is ${\text{CuCl}}_{2}$.

The amount of $\text{NaOH}$ left over is $\text{11 g}$.

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Michael Share
Nov 2, 2014

No chemical reaction occurs here. All the ions are soluble and no new substances form so it is not relevant to write a chemical equation or perform any calculations.

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