Explanation:
It is known that one mole of chromium or molar mass of chromium is 51.99 g/mol.
It is given that number of moles is 11.9 moles.
Therefore, calculate the mass of chromium in grams as follows.
No. of moles =
mass in grams = No. of moles × Molar mass
= 11.9 moles × 51.99 g/mol
= 618.68 g
Thus, we can conclude that there are 618.68 g in 11.9 moles of chromium.
The question is incomplete, here is the complete question.
When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from a mixture of 3.0g Ag and 3.0g .
Answer : The mass of silver sulfide is produced from a mixture is 3.44 grams.
Explanation : Given,
Mass of Ag = 3.0 g
Mass of = 3.0 g
Molar mass of Ag = 107.8 g/mole
Molar mass of = 256 g/mole
Molar mass of = 247.8 g/mole
First we have to calculate the moles of Ag and .
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 16 mole of react with 1 mole of
So, 0.0278 moles of react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 16 mole of react to give 8 mole of
So, 0.0278 moles of react to give
moles of
Now we have to calculate the mass of
Therefore, the mass of silver sulfide is produced from a mixture is 3.44 grams.