Question

How many milliliters of 2.19 M H2SO4 are required to react with 4.75 g of solid containing 21.6 wt% Ba(NO3)2 if the reaction is Ba2+ + SO42- → BaSO4(s)? x mL

Answer 1

1.7927 mL

Step-by-step explanation:

The mass of solid taken = 4.75 g

This solid contains 21.6 wt%
`Ba(NO_3)_2`, thus,

Mass of
`Ba(NO_3)_2` =
`\frac {21.6}{100}* 4.75\ g` = 1.026 g

Molar mass of
`Ba(NO_3)_2` = 261.337 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (1.026\ g)/(261.337\ g/mol)`

`Moles= 0.003926\ mol`

Considering the reaction as:

`Ba(NO_3)_2+H_2SO_4\rightarrow BaSO_4+2HNO_3`

1 moles of
`Ba(NO_3)_2` react with 1 mole of
`H_2SO_4`

Thus,

0.003926 mole of
`Ba(NO_3)_2` react with 0.003926 mole of
`H_2SO_4`

Moles of
`H_2SO_4` = 0.003926 mole

Also, considering:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

Molarity = 2.19 M

So,

`2.19=(0.003926)/(Volume\ of\ the\ solution(L))`

Volume = 0.0017927 L

Also, 1 L = 1000 mL

So, volume = 1.7927 mL