Answer:
38g of Cr
Explanation:
Step 1:
The balanced equation for the reaction:
2K + CrBr2 —> 2KBr + Cr
Step 2:
Determination of the masses of K and CrBr2 that reacted and the mass of Cr produced from the balanced equation.
This is illustrated below:
Molar mass of K = 39g/mol
Mass of K from the balanced equation = 2 x 39 = 78g
Molar Mass of CrBr2 = 52 + (80x2) = 212g
Mass of CrBr2 from the balanced equation = 1 x 212 = 212g
Molar Mass of Cr = 52g/mol
Mass of Cr from the balanced equation = 1 x 52 = 52g
From the balanced equation above,
78g of K reacted with 212g of CrBr2 to produce 52g of Cr.
Step 3:
Determination of the limiting reactant.
This is illustrated below:
From the balanced equation above,
78g of K reacted with 212g of CrBr2.
Therefore, 57g of K will react with = (57 x 212)/78 = 154.92g of CrBr2.
From the above calculation, we can see that a lesser mass (i.e 154.92g) than what was given ( i.e 199g) of CrBr2 is needed to react completely with 57g of K. Therefore, K is the limiting reactant and CrBr2 is the excess reactant.
Step 4:
Determination of the mass of Cr produced by the reaction.
In this case, the limiting reactant will be use because it will give the maximum yield of Cr as all of it is used up in the reaction process. The limiting reactant is K and the mass of Cr produced is obtained as follow:
From the balanced equation above,
78g of K reacted to produce 52g of Cr.
Therefore, 57g of K will produce = (57 x 52)/78 = 38g of Cr.
Therefore, 38g of Cr is produced from the reaction.
