Question

The maximum amount of BaSO4 (in terms of moles) precipitated on mixing equal volumes one litre each of BaCl2 (0.5M) with H2SO4 (1M) WILL CORRESPOND to
1)7kg
2)21kg
​3)14kg
4)28kg

Answer 1

Answer: 0.11669 kg of barium sulfate will be produced.

Explanation:

$[BaCl_2]=0.5 M,[H_2SO_4]=1 M$

$BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl$

Number of moles $BaCl_2$ :

$0.5 M=\frac{\text{moles of } BaCl_2}{1 L}$

Number of moles $BaCl_2$ : 0.5 moles

Number of moles $H_2SO_4$ :

$1 M=\frac{\text{moles of }H_2SO_4}{1 L}$

Number of moles $H_2SO_4$ : 1 moles

According to reaction 1 mole of $BaCl_2$ react with 1 mole of $H_2SO_4$ to give 1 mole of $BaSO_4$.

Then 0.5 moles of $BaCl_2$ react with 0.5 moles of $H_2SO_4$ to give 0.5 moles of $BaSO_4$.

0.5 moles of $BaSO_4$ is present in 2 L of the solution.

Mass of the compound = Number of moles × Molar mass of the compound

Mass of $BaSO_4$= 0.5 moles × 233.38 g/mol=116.69 g=0.11669 kg (1000g =1kg)

0.11669 kg of barium sulfate will be produced.

Answer 2

Answer:

1 4 kg

Explanation:

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is correct for all