Calculate the mass of SbI3 formed when 1.20 g of Sb react with 2.40 g of I2, according to the equation,

2Sb(s) + 3I2(s) ---> 2SbI3(s)

Solution:

To begin this problem we need to build the table.

In this table I decided to start with Sb.

(mol reactant Sb)o

(mol reactant I2)required

(mol reactant I2)o

Conclusion

 

 

 

 

To complete the table we must calculate the mols of Sb and I2 initially,

Entering these values into the table we have,

(mol reactant Sb)o

(mol reactant I2)required

(mol reactant I2)o

Conclusion

9.84 x 10-3

 

9.45 x 10-3

 

Now we need to calculate the mols of I2 required to completely react with 9.84 x 10-3 mols Sb.

So the table now looks like;

(mol reactant Sb)o

(mol reactant I2)required

(mol reactant I2)o

Conclusion

9.84 x 10-3

1.48 x 10-2

9.45 x 10-3

 

 

The next step is determine the limiting reagent.

Looking at the the mols of I2 required and comparing to the mols of I2 initially we can determine whether I2 is the limiting reagent or in excess.

(mol reactant Sb)o

(mol reactant I2)required

(mol reactant I2)o

Conclusion

9.84 x 10-3

1.48 x 10-2

9.45 x 10-3

 

In this case since the mols of I2 required is greater than the mols of I2 initially, we do not have enough I2 to react with all the Sb, therefore I2 is the limiting reagent.

(mol reactant Sb)o

(mol reactant I2)required

(mol reactant I2)o

Conclusion

9.84 x 10-3

1.48 x 10-2

9.45 x 10-3

I2 is limiting

Sb is excess

Click here to check the calculation of the mass SbI3 formed.