Question

Consider the following precipitation reaction: Ba(NO3)2(aq) + K2SO4(aq) → BaSO4(s) + 2 KNO3(aq) If one reacts excess solution of Ba(NO3)2(aq) with 255 mL of a 0.103 M solution of K2SO4(aq), then what mass of BaSO4 will be produced?

Answer 1

To find the mass of BaSO4 produced from a precipitation reaction between Ba(NO3)2 and K2SO4, one calculates the moles of K2SO4, then uses a 1:1 stoichiometry and the molar mass of BaSO4 to find the precipitate's mass. In this case, 6.128 g of BaSO4 will be produced.

Step-by-step explanation:

Calculating the Mass of BaSO4 Precipitate

To calculate the mass of BaSO4 produced, we must first calculate the number of moles of K2SO4 present in the solution. As we have a 0.103 M solution, and the volume of the solution is 255 mL, which is 0.255 L, we can find the moles of K2SO4 by multiplying the concentration by the volume in liters:

Moles of K2SO4

= 0.103 mol/L × 0.255 L = 0.026265 mol

Since the reaction between Ba(NO3)2 and K2SO4 yields BaSO4 in a 1:1 stoichiometry, the moles of BaSO4 formed will be the same as the moles of K2SO4 reacted, assuming Ba(NO3)2 is in excess.

Now, we use the molar mass of BaSO4 (233.39 g/mol) to find the mass of BaSO4:

Mass of BaSO4

= Moles of BaSO4 × Molar mass of BaSO4 = 0.026265 mol × 233.39 g/mol =

6.128 g

Therefore, 6.128 g of BaSO4 will be produced when excess Ba(NO3)2 reacts with 255 mL of 0.103 M K2SO4 solution.