Question

How many grams of barium sulfate are formed from reaction of 10.0 mL of 0.250 M (NH4)2SO4 And excess BaCl2?

How many grams of barium sulfate are formed from reaction of 10.0 mL of 0.250 M (NH4)2SO4 and excess BaCl2?

Answer 1

Balanced chemical equation is:

(NH4)2SO4 + BaCl2 —> BaSO4 + 2 NH4Cl

lets calculate the mol of (NH4)2SO4

volume , V = 10 mL

= 1*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.25*1*10^-2

= 2.5*10^-3 mol

According to balanced equation

mol of BaSO4 reacted = (1/1)* moles of (NH4)2SO4

= (1/1)*2.5*10^-3

= 2.5*10^-3 mol

This is number of moles of BaSO4

Molar mass of BaSO4,

MM = 1*MM(Ba) + 1*MM(S) + 4*MM(O)

= 1*137.3 + 1*32.07 + 4*16.0

= 233.37 g/mol

use:

mass of BaSO4,

m = number of mol * molar mass

= 2.5*10^-3 mol * 2.334*10^2 g/mol

= 0.5834 g

Answer: 0.583 g