Question

Starting with 1.5052g of BaCl2•2H2O and excessH2SO4, how many grams of BaSO4 can be formed?

Answer 1

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 2

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.