# ONE QUESTION ON STOICHIOMETRY.

TRY TO SOLVE THIS PROBLEM ON STOICHIOMETRY.

QUESTION

How many grams of solid barium sulphate is formed, when 25ml of 0.160M of BaCl2 reacts with 68ml of 0.055M of Na2SO4 ?

(A). 87g

(B). 8.7g

(C). 0.87g

(D). None

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May 19, 2020 6:24 pm

Thomas
C

May 18, 2020 11:48 am

D

May 18, 2020 11:48 am

D

May 17, 2020 3:00 pm

0.83

May 16, 2020 10:38 pm

Benny, its 0.83g

May 16, 2020 1:47 pm

De and is D none

May 14, 2020 11:38 pm

Josh

Ans= D
Mass of BaSO4 in grams = 0.932g
Molarity or concentration of BaSO4 = 0.043mol/L

May 14, 2020 8:25 pm

Marycynthia
0.932 g of Baso4

May 14, 2020 8:20 pm

D. None

May 14, 2020 5:21 pm

Orisakwe Emmanuel
Ans is None

May 14, 2020 5:20 pm

Orisakwe Emmanuel

Equation for the reaction..

Bacl2+Na2SO4=Na2CL2 + BaSO4
Number of moles: 1 1 = 1 1
Mole ratio: 1:1:1:1

Number of moles formed in Bacl2 and NaSO4

For Bacl2

Mole=0.160mx0.025l
=4*10^-3
For NaSO4

Mole=0.068*0.055
=3.74*10^-3
Or 4*10^-3

In other words Bacl2 reacted with NaSO4 To form BaSO4

So mole of BaSO4= 4*10^-3
Grams that was formed in BaSO4…..

BaSO4 volume formed= total volume of Bacl2 + NaSO4
= 0.093
Conc= 4*10^-3/0.093
Ans= 0.043g

May 14, 2020 5:15 pm

D

The answer is 0.934g of BaSo4

May 14, 2020 4:46 pm

Daniel
C.0.87g of BaSO4

May 14, 2020 4:32 pm

From Benitha..
D previous ans.I gave was wrong..
0.932g of Baso4 is correct..
D is d correct ans..

May 14, 2020 4:11 pm

From Benitha
0.832g..
None

May 14, 2020 3:42 pm

D None
Precious

May 14, 2020 2:06 pm

0.87g

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