Question

CanvasX &Question 16Consider the reaction: 4Na(s) + O2(8) → 2Na2O(s)

Answer 1

The mass of Na2O in grams is 123.49 grams

Step-by-step explanation:

Given information

The mass of sodium is 91 grams

The mass of oxygen is 34.0 grams

To find the amount of Na2O produced, follow the steps below

Step 1: Write the balanced equation of the reaction

`\text{ 4Na}_((s))\text{ + O}_(2(g))\rightarrow\text{ 2Na}_2O_((S))`

Step 2: Find the number of moles of sodium and oxygen using the formula below

`\text{ Mole = }\frac{\text{ mass}}{molar\text{ mass}}`

Recall, that the molar mass of Na is 22.989 g/mol, and the molar mass of oxygen atom is 32 g/mol

For Na

`\begin{gathered} \text{ Mole = }\frac{mass}{molar\text{ mass}} \\ \text{ Mole = }(91.6)/(22.989) \\ \text{ Mole = 3.985 moles} \end{gathered}`

For O2

`\begin{gathered} \text{ Mole = }\frac{mass}{molar\text{ mass}} \\ \text{ } \\ \text{ Mole = }(34)/(32) \\ \text{ Mole = 1.0625 moles} \end{gathered}`

Step 3: Find the limiting reactant of the reaction

The limiting reactant is the reactant that has the least number of moles after dividing it by the coefficient of the reactant

From the reaction above, you will see that 4 moles of sodium react with one mole of oxygen to give 2 moles of sodium oxide.

For Na

`\begin{gathered} \text{ Na = }\frac{\text{ number of moles}}{\text{ coefficient}} \\ \text{ Na = }(3.985)/(4) \\ \text{ Na = 0.99625 mole/wt} \\ \\ \text{ For O}_2 \\ \text{ O}_2\text{ = }(1.0625)/(1) \\ \text{ O}_2\text{ = 1.0625 moles/wt} \end{gathered}`

From the above calculations, you will see that Na has the least number of moles, hence, Na is the limiting reactant

Step 4: Find the number of moles of Na2O

The Na2O can be determined using a stoichiometric ratio

Let x represents the number of moles

`\begin{gathered} \text{ 4 moles of Na }\rightarrow\text{ 2 moles of Na}_2O \\ 3.985\text{ moles of Na }\rightarrow\text{ x moles of Na}_2O \\ \text{ cross multiply} \\ \text{ 4 moles of Na }*\text{ x moles of Na}_2O\text{ = 2 moles of Na}_2O*3.985\text{ moles of Na} \\ \text{ x moles of Na}_2O\text{ = }\frac{2\cancel{moles\text{ of Na}_2}O*3.985\cancel{moles\text{ of Na}}}{4\cancel{moles\text{ of Na}}} \\ \text{ x moles of Na}_2O\text{ = }\frac{2*\text{ 3.985}}{4} \\ \text{ X moles of Na}_2O\text{ = 1.9925 moles} \end{gathered}`

Hence, the number of moles of Na2O is 1.9925 moles

Step 5: Find the mass of Na2O

`\begin{gathered} \text{ Mole = }\frac{\text{ mass}}{\text{ molar mass}} \\ \end{gathered}`

Recall, that the molar mass of Na2O is 61.9789 g/mol

`\begin{gathered} \text{ 1.9925 = }\frac{\text{ mass}}{61.978} \\ \text{ Cross multiply} \\ \text{ Mass = 1.9925}*61.978 \\ \text{ Mass = 123.49 grams} \end{gathered}`

Hence, the mass of Na2O in grams is 123.49 grams