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

Answer:

1.7927 mL

Explanation:

The mass of solid taken = 4.75 g

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

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

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

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.026\ g}{261.337\ g/mol}$

$Moles= 0.003926\ mol$

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:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Molarity = 2.19 M

So,

$2.19=\frac{0.003926}{Volume\ of\ the\ solution(L)}$

Volume = 0.0017927 L

Also, 1 L = 1000 mL

So, volume = 1.7927 mL