Verified Answer :
1.5052g BaCl2.2H2O => 1.5052g / 274.25 g/mol = 0.0054884 mol
=> 0.0054884 mol Ba
This means that at most 0.0054884 mol BaSO4 can form since Ba is the limiting reagent.
0.0054884 mol BaSO4 => 0.0054884 mol * 233.39 g/mol = 1.2809 g BaSO4
Answer: The mass of barium sulfate produced is 1.45 grams.
Explanation:
To calculate the number of moles, we use the equation:
\(\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\) .....(1)
Given mass of \(BaCl_2.2H_2O\) = 1.5052 g
Molar mass of \(BaCl_2.2H_2O\) = 244.26 g/mol
Putting values in equation 1, we get:
\(\text{Moles of }BaCl_2.2H_2O=\frac{1.5052g}{244.26g/mol}=0.0062mol\)
The chemical equation for the reaction of barium chloride and sulfuric acid follows:
\(BaCl_2.2H_2O+H_2SO_4\rightarrow BaSO_4+2HCl+2H_2O\)
As, sulfuric acid is present in excess, it is considered as an excess reagent.
Thus, \(BaCl_2.2H_2O\) is considered as a limiting reagent because it limits the formation of product
By Stoichiometry of the reaction:
1 mole of \(BaCl_2.2H_2O\) produces 1 mole of barium sulfate
So, 0.0062 moles of \(BaCl_2.2H_2O\) will produce = \(\frac{1}{1}\times 0.0062=0.0062\) moles of barium sulfate
Now, calculating the mass of barium sulfate by using equation 1:
Molar mass of barium sulfate = 233.4 g/mol
Moles of barium sulfate = 0.0062 moles
Putting values in equation 1, we get:
\(0.0062mol=\frac{\text{Mass of barium sulfate}}{233.4g/mol}\\\\\text{Mass of barium sulfate}=(0.0062mol\times 233.4g/mol)=1.45g\)
Hence, the mass of barium sulfate produced is 1.45 grams.