Respuesta :

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:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

Given mass of [tex]BaCl_2.2H_2O[/tex] = 1.5052 g

Molar mass of [tex]BaCl_2.2H_2O[/tex] = 244.26 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }BaCl_2.2H_2O=\frac{1.5052g}{244.26g/mol}=0.0062mol[/tex]

The chemical equation for the reaction of barium chloride and sulfuric acid follows:

[tex]BaCl_2.2H_2O+H_2SO_4\rightarrow BaSO_4+2HCl+2H_2O[/tex]

As, sulfuric acid is present in excess, it is considered as an excess reagent.

Thus, [tex]BaCl_2.2H_2O[/tex] is considered as a limiting reagent because it limits the formation of product

By Stoichiometry of the reaction:

1 mole of [tex]BaCl_2.2H_2O[/tex] produces 1 mole of barium sulfate

So, 0.0062 moles of [tex]BaCl_2.2H_2O[/tex] will produce = [tex]\frac{1}{1}\times 0.0062=0.0062[/tex] 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:

[tex]0.0062mol=\frac{\text{Mass of barium sulfate}}{233.4g/mol}\\\\\text{Mass of barium sulfate}=(0.0062mol\times 233.4g/mol)=1.45g[/tex]

Hence, the mass of barium sulfate produced is 1.45 grams.

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