How do I do this problem?

Question

How many grams of Cdl can be theoretically produced by reaction between 10-g od Cd and 20-g of I2?
MM(Cd) = 112g/mol
MM(I2) = 254 g/mol
MM(Cdl) = 239 g/mol

the answer is 21, but how do you get that?
thanks

Answer 1

assuming 2mol Cd + 1mol I2 ------> 2mol CdI
first calculate the amount of mols of each chemical:
10g Cd* 1mol Cd/112g= 0.089 mol Cd
20g I2* 1mol I2/254g= 0.079mol I2

since we need 2 mols of Cd for every 1 mol of I2 to produce 1 mol of product, multiply the amount of mols of I2 by 2 to see if it equals the amount of mols of Cd (or you can divide the mols of Cd by 2 and see if it equals the amounf of mols of I2):

0.079mol I2*2=0.159 does not equal 0.089, in fact we have more I2 than we need since each mol reacts with 2 mols of Cd, therefore Cd is the limiting reactant. THE AMOUNT OF PRODUCT IN MOLs IS PROPORTIONAL TO THE AMOUNT OF LIMITING REAGENT:

____mol CdI= .089mol Cd ; 2mol Cd= 2mol CdI

Since 2mol Cd yields 2 mol CdI, it is a 1:1 relationship: 1 mol Cd makes 1 mol CdI (when reacted with half a mol of I2). Thus the amount of mols of CdI produced is 0.089, the same as the amount of Cd reacted. To find the mass produced multiply by molar mass:

0.089 mol CdI* 239g CdI/mol CdI = 21.2g CdI

Answer 2

you must have since Cd I seems valence 1

2Cd + I2 = 2 CdI

1O g of Cd correspond to 0.089 mole of Cd (10/112)
20g of I2 correspond to 20/254 = 0.0787 moles of I2

You need 2 moles of Cd for 1 mole of I2

if you have 0.089 mole of Cd , the Cd is the limiting factor since it will react with 0.0393 mol of I2

So, you will have Iodine in excess and obtain 0.089 moles of CdI

with a weight 0.089 *239 = 21 g of CdI

Answer 3

find the number of moles of the limiting reagent and then multiply it by the weight of the CdI.

10/112 = 0.089285714 moles Cd
2*(20/254) = 0.157480315 moles I

Cd is the limiting reagent so...

0.089285714 moles CdI * 239 g/mole = 21.339285714 g CdI