How many atom of oxygen are in 17.7g of potassium promangenate?

Answer 1

Assuming that the questioner meant to write "permanganate" instead of "promangenate", the answer may be found as follows:

The formula of potassium permanganate is KMnO4, showing four atoms of oxygen per formula unit. The formula unit mass of KMnO4 is 158.03; therefore, 17.7 g of potassium permanganate contains [4(17.7)/158.03](Avogadro's Number) or 2.70 X 1023 atoms of oxygen, to the justified number of significant digits.

Answer 2

To find the number of oxygen atoms in 17.7g of potassium permanganate (KMnO4), we need to use Avogadro's number and the molar mass of KMnO4. The molar mass of KMnO4 is 158.034 g/mol. Thus, 17.7g of KMnO4 is equal to 0.112 moles. In one mole of KMnO4, there are 4 oxygen atoms. Therefore, in 17.7g of KMnO4, there are 0.112 moles x 4 = 0.448 moles of oxygen atoms. Finally, using Avogadro's number (6.022 x 10^23 atoms/mol), we find that there are approximately 2.7 x 10^23 oxygen atoms in 17.7g of KMnO4.