Question

You have seven closed containers, each with equal masses of chlorine gas (CI₂). You add 10.09 of sodium to the first sample, 20.0 g of sodium to the second sample, and so on (adding 70.0 g of sodium to the seventh sample). Sodium and chlorine react to form sodium chloride according to the equation

$\mathbf{2}\mathit{N}\mathit{a}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{I}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathit{N}\mathit{a}\mathit{C}\mathit{I}\mathbf{\left(}\mathit{s}\mathbf{\right)}$

After each reaction is complete, you collect and measure the amount of sodium chloride formed.

A graph of your results is shown below.

Answer the following questions:
a. Explain the shape of the graph.
b. Calculate the mass of NaCI formed when 20.0 g of sodium is used.
C. Calculate the mass of, (CI₂) in each container.
d. Calculate the mass of NaCIformed when 50.0 g of sodium is used.

Short answer

1. The shape of the graph explains that the amount of (CI)determines the amount of NaCI.
2. The mass ofNaCI formed when 20.0 g of sodium used is50.84g.
3. The mass of (CI₂), in each container is 60.66g.
4. The mass ofNaCI formed when 50.0 g of sodium used is 125g.

Step-by-step solution

Definition

The molar mass of a material is defined as the mass of the substance contained in one mole of the substance, and it isequal to the total mass of the component atoms. It may be stated mathematically as follows:
Molar mass =$\frac{Weighting}{numberofmoles}$

Explanation of the graph

Each of the seven closed containers is filled with the same amount of (CI₂) and various amounts of Na.

We can see from the graph that adding 10 g, 20 g, and 30 g Na produces different amounts of NaCI, whereas adding 40, 50, 60, and 70 g Na produces the same amount of NaCI, implying that the limiting reactant in cases where 40, 50, 60, and 70 g Na is added is (CI₂), whereas the limiting reactant in the first three cases is Na.

a)

When(CI₂) becomes the limiting reactant, adding more Na in a container gives more product and the graph has a positive slope, but when (CI₂) becomes the limiting reactant, adding more Na doesn't produce a greater mass ofsince the amount of (CI₂) determines the amount of NaCI

Step 3: Calculation for the mass of NaCI formed when 20.0 g of sodium is used

b)

The balanced equation,

$$\begin{gathered} 2Na\left(s\right)+C{I}_2\left(g\right)\to 2NaCI\left(s\right) \\ M\left(Na\right)=22.99\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right. \\ n\left(Na\right)=20gNa\times \frac{1mol}{22.99gNa} \\ =0.87mol \\ M\left(NaCI\right)=22.99+35.45 \\ =58.44\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right. \\ m\left(NaCI\right)=0.87molNaCI\times \frac{58.44g}{1molNaCI} \\ =50.84g \end{gathered}$$

Therefore the mass of NaCI is50.84g.

Step 4: Calculation for the mass of (CI2)

c)

We can determine the mass of (CI₂) when 50 g Na is added, for example, because (CI₂) is the limiting reactant.

We can see from the graph that m(NaCI) creates 100 g when 50 g Na is added (we know this since m(NaCI) equals 50.84 g when 10 g Na is added, hence one square in a mass of NaCI is 125 g).

$$\begin{gathered} n\left(NaCI\right)=100gNaCI\times \frac{1mol}{58.44gNaCI} \\ =1.711mol \end{gathered}$$

Thus the mol ratio is,

$$\begin{gathered} \frac{1molC{I}_2}{2molNaCI} \\ n\left(C{I}_2\right)=1.711molNaCI\left(\frac{1molC{I}_2}{2molNaCI}\right) \\ =0.855molC{I}_2 \end{gathered}$$

Let’s convert this to mass,

$$\begin{gathered} m\left(C{I}_2\right)=0.8551molC{I}_2\left(\frac{70.9gC{I}_2}{1molNaCI}\right) \\ =60.66gC{I}_2 \end{gathered}$$

Step 5: Calculation for the mass of NaCI formed when 50.0 g of sodium is used

d)

We can see from the graph that when 50 g Na is added, m(NaCI) creates 100 g (we know this because when 10 g Na is added, m(NaCI) produces 50.84 g, hence one square in a mass of NaCI is 125 g).