Question

If you have 35.0g of Na and 100.0g of Cl2, what is the maximum amount of NaCl that you can produce?

Answer 1

89.0 g

Step-by-step explanation

Given information:

Equation of the reaction: 2Na + Cl₂ → 2NaCl₂

Reacting mass of Na = 35.0 g

Reacting mass of Cl₂ = 100.0 g

What to find:

The maximum amount of NaCl that can be produce from 35.0g of Na and 100.0g of Cl₂.

Step-by-step solution:

From the given balanced chemical equation for the reaction; 2 moles of Na reacts with 1 mole of Cl₂ to produce 2 moles of NaCl₂.

The next step is to convert the given masses to moles

From the Periodic Table:

Molar mass of Na = 22.989769 g/mol

Molar mass of Cl₂ = 35.453 g/mol

The formula to calculate mole is:

`\text{mole }=\frac{\text{Reacting mass}}{\text{Molar mass}}`

Therefore,

`\begin{gathered} \text{moles of Na }=\frac{35.0\text{ g}}{22.989769g\text{/mol}}=1.52\text{ moles} \\ \\ \text{moles of Cl}_(2)\text{ }=\frac{100.0\text{ g}}{35.453g\text{/mol}}=2.82\text{ moles} \end{gathered}`

The next step is to determine the limiting reactant.

To get that, divide the moles of Na and Cl₂ with their respective coefficients in the given equation:

For Na = 1.52/2 = 0.75

For Cl₂ = 2.82/1 = 2.82

Hence, Na is the limiting reactant.

Now you can compare the mass of the mass of Na to the mass of NaCl₂ produced as follows:

`\begin{gathered} (2*22.989769)g\text{ Na produced }2(22.989769+35.453)g\text{ NaCl}_(2) \\ i\mathrm{}e\text{ }45.979538\text{ g Na produced 116.885538 g NaCl}_(2) \\ So,\text{ 35.0 g Na will produce x g NaCl}_(2) \\ \text{x g NaCl}_(2)=\frac{\text{35.0 g Na}*\text{116.885538 g NaCl}_(2)}{\text{ }45.979538\text{ g Na}}=88.97422654g\text{ NaCl}_(2) \\ x\text{ g NaCl}_(2)\text{ }\approx89.0g\text{ NaCl}_(2) \end{gathered}`

The maximum amount of NaCl that can be produce from 35.0g of Na and 100.0g of Cl₂ is 89.0 g