Question

How many molecules of sodium oxide will be created if 187 grams of oxygen reacts with excess sodium? 4na o2 -> 2na2o

Answer 1

Molecules of sodium oxide will be created if 187 grams of oxygen reacts with excess sodium will be 7.04 × 10 ²⁴ molecules.

Sodium oxide is Na₂O which is formed due to the reaction of sodium and oxygen.

Given,

Balanced chemical equation is :

         4 Na(s) + O₂ (g) → 2Na₂O (s)

Now, the Mole ratio of O₂ and Na₂O is 1 mol O₂: 2 mol Na₂O

Molar mass O₂ = 32.00 g/mol

Now, let's find out the Number of moles of O₂ present in 187 grams of oxygen,

Number of moles of O₂ = Given mass in grams / molar mass of O₂

Number of moles of O₂ = 187 g / 32.00 g/mol

Number of moles of O₂ = 5.844 mol

Since, we know the Proportion is 1 mol O₂ : 2 mol Na₂O

Now, Number of moles of Na₂O required is,

2 mol Na₂O / 1 mol O₂ = 5.844 mol O₂ / x

x = 5.844 mol O₂ (2 mol Na₂O / 1 mol O₂)

x = 11.688 mol Na₂O

To calculate the Number of molecules of sodium oxide,

Number of molecules = Number of moles × Avogadro's number

Number of molecules = 11.688 mol × 6.022 × 10²³ molecules/mol =

Number of molecules = 7.04 × 10 ²⁴ molecules

Hence, the Number of molecules of sodium oxide required is 7.04 × 10²⁴ molecules.                          

Learn more about Sodium oxide here,

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