Suppose 13.6 g of barium nitrate is dissolved in 300. ml of a 0.40m aqueous solution of sodium chromate. calculate the final molarity of barium cation in the solution. you can assume the volume of the solution doesn't change when the barium nitrate is dissolved in it. be sure your answer has the correct number of significant .

The molarity of barium cation in the solution is 0.173 M

Explanation:

Step 1: The balanced equation

Ba(NO3)2(aq) + Na2CrO4 (aq) → BaCrO4(s) + 2NaNO3(aq)

Step 2: Data given

Mass of Barium nitrate = 13.6 grams

Volume of 0.40M sodium chromate = 300 mL

Step 3: Calculate moles of Ba(NO3)2

Moles = mass / molar mass

Moles = 13.6 grams / 261.34 g/mol

Moles = 0.052 moles

Step 4: Calculate moles of Na2CrO4

Moles = Molarity * Volume

Moles Na2CrO4 = 0.40 * 0.3L

Moles Na2CrO4 = 0.12 moles

Step 5: Calculate limiting reactant

Na2CrO4 is in excess so all of Ba(NO3)2 will be consumed and reacts to form BaCrO4(s) in the form Ba2+

Step 6: Calculate moles of Ba2+

n(Ba2+)=n(BaCrO4) =n(Ba(NO3)2 = 0.0520 moles

Step 7: Calculate molarity of Ba2+

C=n/v so C(Ba2+)=0.0520/0.300 = 0.173 M

The molarity of barium cation in the solution is 0.173 M

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