Respuesta :
Answer:
Sulfuric acid is limiting reagent. The formula of limiting reagent is [tex]H_2SO_4[/tex].
Mass of barium sulfate formed is 71.62 g
Excess Mass of barium sulfate remained is 4 g
Explanation:
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
For barium hydroxide
Mass of barium hydroxide = 56.6 g
Molar mass of barium hydroxide = 171.34 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{56.6\ g}{171.34\ g/mol}[/tex]
[tex]Moles\ of\ barium\ hydroxide= 0.3303\ mol[/tex]
For sulfuric acid
Mass of sulfuric acid = 30.1 g
Molar mass of sulfuric acid = 98.079 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{30.1\ g}{98.079\ g/mol}[/tex]
[tex]Moles\ of\ sulfuric\ acid= 0.3069\ mol[/tex]
According to the given reaction:
[tex]Ba(OH)_2+H_2SO_4\rightarrow BaSO_4+2H_2O[/tex]
1 mole of sulfuric acid reacts with 1 mole of barium hydroxide
So,
0.3069 mole of sulfuric acid reacts with 0.3069 moles of barium hydroxide
Moles of barium hydroxide = 0.3069 moles
Available moles of barium hydroxide = 0.3303 moles
Limiting reagent is the one which is present in small amount. Thus, sulfuric acid is limiting reagent. The formula of limiting reagent is [tex]H_2SO_4[/tex].
The formation of the product is governed by the limiting reagent. So,
1 mole of sulfuric acid produces 1 mole of barium sulfate
So,
0.3069 mole of sulfuric acid produces 0.3069 mole of barium sulfate
Moles of barium sulfate = 0.3069 mole
Molar mass of barium sulfate = 233.38 g/mol
Mass of barium sulfate = Moles × Molar mass = 0.3069 × 233.38 g = 71.62 g
Excess reactant = barium hydroxide = 0.3303 moles - 0.3069 moles = 0.0234 moles
Molar mass of barium hydroxide = 171.34 g/mol
Mass of barium sulfate = Moles × Molar mass = 0.0234 × 171.34 g = 4 g
