For the following reaction, 67.1 grams of barium hydroxide are allowed to react with 36.3 grams of sulfuric acid:

\[ \text{barium hydroxide (aq)} + \text{sulfuric acid (aq)} \rightarrow \text{barium sulfate (s)} + \text{water (l)} \]

What is the maximum amount of barium sulfate that can be formed?

Question

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Answer :

kobenhavn kobenhavn · Answer 1 · 2023-11-03T18:51:08-04:00

Answer: 86.2 g of [tex]BaSO_4[/tex] will be produced from the given masses of both reactants.

Explanation:

To calculate the moles, we use the equation:

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

a) moles of [tex]Ba(OH)_2[/tex]

[tex]\text{Number of moles}=\frac{67.1g}{171g/mol}=0.39moles[/tex]

b) moles of [tex]H_2SO_4[/tex]

[tex]\text{Number of moles}=\frac{36.3g}{98g/mol}=0.37moles[/tex]

[tex]Ba(OH)_2+H_2SO_4\rightarrow BaSO_4+2H_2O[/tex]

According to stoichiometry :

1 moles of [tex]H_2SO_4[/tex] require 1 mole of [tex]Ba(OH)_2[/tex]

Thus 0.37 moles of [tex]H_2SO_4[/tex] require=[tex]\frac{1}{1}\times 0.37=0.37moles[/tex] of [tex]Ba(OH)_2[/tex]

Thus [tex]H_2SO_4[/tex] is the limiting reagent as it limits the formation of product.

As 1 mole of [tex]H_2SO_4[/tex] give = 1 mole of [texBaSO_4[/tex]

Thus 0.37 moles of [tex]H_2SO_4[/tex] give =[tex]\frac{1}{1}\times 0.37=0.37moles[/tex] of [tex]BaSO_4[/tex]

Mass of [tex]BaSO_4=moles\times {\text {Molar mass}}=0.37moles\times 233g/mol=86.2g[/tex]

Thus 86.2 g of [tex]BaSO_4[/tex] will be produced from the given masses of both reactants.