PLEASE HELPP!! Consider the reaction below.
2 Na + O2 → Na2O2
If 10.0 g of sodium metal reacts with excess oxygen gas how many grams of sodium oxide will be produced?
Show your work below. Make sure to show ALL units and each step of the process:

 The correct answer is Option B.

Explanation:

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}Number of moles=Molar massGiven mass   (1)

For N_2N2 :

Given mass of nitrogen gas = 10 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

\text{Moles of iron oxide}=\frac{10g}{28g/mol}=0.357molMoles of iron oxide=28g/mol10g=0.357mol

The given chemical reaction follows:

N_2+O_2\rightarrow 2NON2+O2→2NO

As, oxygen gas is present in excess. Thus, it is considered as an excess reagent and nitrogen is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

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