Answer:
35.42g
Explanation:
Step 1:
The balanced equation for the reaction
BaCl2(aq) + K2CrO4(aq) → BaCrO4(s) + 2KCl(aq)
Step 2:
Determination of the limiting reactant.
It is important to determine which of the reactant is limiting the reaction as the limiting reactant is used to determine the maximum yield of the reaction. The limiting reactant can be determined as follow:
From the balanced equation above,
1 mole of BaCl2 reacted with 1 mole of K2CrO4.
Therefore, 0.7 mole of BaCl2 will also react with 0.7 mol of K2CrO4.
From the above illustration, we can see that it requires a higher amount of K2CrO4 to react with 0.7 mol of BaCl2. This simply means that K2CrO4 is the limiting reactant.
Step 3:
Determination of the number of mole of BaCrO4 produced from the reaction.
The limiting reactant is used in this case.
From the balanced equation above,
1 mole of K2CrO4 produced 1 mole of BaCrO4.
Therefore, 0.14 mole of K2CrO4 will also produce 0.14 mole of BaCrO4.
Step 4:
Converting 0.14 mole of BaCrO4 to grams.
This is illustrated below:
Molar Mass of BaCrO4 = 137 + 52 + (16x4) = 137 + 52 + 64 = 253g/mol
Number of mole BaCrO4 = 0.14 mole
Mass of BaCrO4 =?
Mass = number of mole x molar Mass
Mass of BaCrO4 = 0.14 x 253
Mass of BaCrO4 = 35.42g
Therefore, 35.42g of BaCrO4 is produced from the reaction.
