Question

Calculate the number of Na, C and O atoms present in 5.3 gram of sodium carbonate ( Na2CO3)​

Answer 1

Answer:

Na - 2

C - 1

O - 3

hope it helps you

Answer 2

Calculation of the number of Na (Sodium), C (Carbon) and O (Oxygen) in Na₂CO₃

Explanation:

Given compound is Na₂CO₃,here we have to calculate the number of Na (Sodium), C (Carbon) and O(Oxygen) with the given amount of sodium carbonate.

Here Given Sodium carbonate is 5.3 gm.

In this amount of sodium carbonate, the number of Sodium, carbon and Oxygen is determined using the following steps-

Calculation:

Step 1 of 3

Here, the mass of sodium carbonate given is = 5.3g

Formula mass  of Na₂CO₃=  (2×23+12+3×16)

so  formula mass of Sodium carbonate = 106

Now 106g of Na₂CO₃= 6.023×10²³ formula units ofNa₂CO₃  

Step 2of 3

5.3g sodium carbonate = (5.3÷106) mol

= 0.05mol

Number of sodium carbonate molecule = = 3.01×10²²

Step 3 of 3

In one formula unit of Na₂CO₃ contains 2 atoms of Na (0.05×6.023×10²³sodium), 1 atom of carbon,3 atoms of Oxygen

Therefore number of Na atoms in 5.3 g of  Na₂CO₃=(2×3.01×10²²)

=6.023×10²²

No of carbon atoms in 5.3 g of Na₂CO₃ =(1×3.01×10²²)

=3.01×10²²

No of Oxygen atoms in 5.3 g of Na₂CO₃ =(3×3.01×10²²)

=9.03×10²²

Answer =Na₂CO₃ contains two atoms of Sodium , 1 atom of Carbon and 3 atoms of Oxygen