One step in the isolation of pure rhodium metal is the precipitation of rhodium (III) hydroxide from a solution containing rhodium (III) sulfate according to the following balanced chemical equation:

 

Rh2(SO4)3 + 6NaOH --> 2Rh(OH)3 + 3Na2SO4

 

If 0.620g of rhodium (III) hydroxide is pro duced, what mass of sodium sulfate is also produced?

 

choices:

 

0.572g

0.930g

0.858g

0.620g

0.381g

 

One step in the isolation of pure rhodium metal is the precipitation of rhodium (III) hydroxide from a solution containing rhodium (III) sulfate according to the following balanced chemical equation:

 

Rh2(SO4)3 + 6NaOH --> 2Rh(OH)3 + 3Na2SO4

 

If 0.620g of rhodium (III) hydroxide is pro duced, what mass of sodium sulfate is also produced?

 

choices:

 

0.572g

0.930g

0.858g

0.620g

0.381g

 

 

You have to use molar ratios for this. So convert the 0.62g to moles, then multiply by the molar ratio of the two products. so n*(3/2).

Then convert back to grams and that will be the mass of sodium sulfate

AHHH I LOVE YOU!!! Answer is: 0.858g