what mass of barium sulphate can be produced when 100.0 ml of a 0.100M solution of barium chloride is mixed with 100.0 ml of a 0.100M solution of Iron(lll) sulphate?
Write the equation and balance it. This is a limiting reagent problem; you know that because amounts for BOTH reactants are given. I solve these by using reagent 1 and ignoring the second and calculating the product. Then repeat the process using reagent 2 and ignoring reagent 1 and calculating the product. You will get two different answers and only one can be correct; the correct one is ALWAYS the smaller one. Here is a worked example of a stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Post your work if you get stuck.
which of them will be the limiting reactant??
The limiting reagent is the one producing the smaller value of the product.
To determine the mass of barium sulfate that can be produced, we need to identify the limiting reactant in the reaction between barium chloride (BaCl2) and iron(III) sulfate (Fe2(SO4)3). The limiting reactant is the one that is completely consumed and determines the amount of product formed.
First, let's write the balanced chemical equation for the reaction between barium chloride and iron(III) sulfate:
3 BaCl2 + Fe2(SO4)3 -> 3 BaSO4 + 2 FeCl3
From the balanced equation, we can see that it takes 3 moles of barium chloride (BaCl2) to react with 1 mole of iron(III) sulfate (Fe2(SO4)3) to produce 3 moles of barium sulfate (BaSO4).
Now, let's calculate the number of moles of barium chloride (BaCl2) and iron(III) sulfate (Fe2(SO4)3) present in the solutions:
Volume of BaCl2 solution = 100.0 mL
Concentration of BaCl2 solution = 0.100 M
Number of moles of BaCl2 = (Volume in liters) x (Concentration)
= (100.0 mL / 1000 mL/L) x 0.100 M
= 0.010 moles
Similarly, for iron(III) sulfate:
Volume of Fe2(SO4)3 solution = 100.0 mL
Concentration of Fe2(SO4)3 solution = 0.100 M
Number of moles of Fe2(SO4)3 = (Volume in liters) x (Concentration)
= (100.0 mL / 1000 mL/L) x 0.100 M
= 0.010 moles
Since the stoichiometric ratio between BaCl2 and Fe2(SO4)3 is 3:1, we can see that both the moles of BaCl2 and Fe2(SO4)3 are the same. Therefore, the limiting reactant is barium chloride (BaCl2).
Now, let's calculate the number of moles of barium sulfate (BaSO4) that can be produced from the limiting reactant:
Number of moles of BaSO4 = 3 x Number of moles of BaCl2
= 3 x 0.010 moles
= 0.030 moles
Finally, let's calculate the mass of barium sulfate using its molar mass:
Molar mass of BaSO4 = 137.33 g/mol (approximately)
Mass of BaSO4 = Number of moles x Molar mass
= 0.030 moles x 137.33 g/mol
= 4.12 grams
Therefore, approximately 4.12 grams of barium sulfate can be produced when 100.0 mL of a 0.100 M solution of barium chloride is mixed with 100.0 mL of a 0.100 M solution of iron(III) sulfate.