How many atoms of carbon are contained in 47.6 g of al2(co3)3? the molar mass of al2(co3)3 is 233.99 g/mol?

The molar mass of Al₂(CO₃)₃is 233.99 g/mol

It means 233.99 grams of Al₂(CO₃)₃ are in one mole

And moles in 47.6 g = 47.6/233.99 = 0.203 moles

Now we know that we have 0.203 moles of Al₂(CO₃)₃

In one mole of Al₂(CO₃)₃, we have 3 moles of carbon

To find the number of moles of carbon in a 0.203 moles of Al₂(CO₃)₃, we must multiply this by the 3 moles of carbon 

0.203 x 3 = 0.609 moles of carbon 

Within one mole of carbon, we have 6.022 x 10²³ atoms. So, to find the amount of atoms within 0.609 moles of carbon, we must multiply the 0.609 moles by 6.022 x 10²³ atoms. 

So, the number of carbon atoms present in 47.6g of Al₂(CO₃)₃ = 0.609 x 6.022 x 10²³ = 3.667 x 10²³ atoms.

explanation:

two years ago, when picking flowers in the mountains back of yosemite valley, i found a book. it was blotted and storm-beaten; all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however; many of the inner pages were well preserved. and although all were more or less stained and torn, whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today; its granite pages have been torn and blurred by the same storms that wasted the castaway book.

two years ago, when picking flowers in the mountains back of yosemite valley; i found a book. it was blotted and storm-beaten, all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however, many of the inner pages were well preserved. and although all were more or less stained and torn; whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today, its granite pages have been torn and blurred by the same storms that wasted the castaway book.