Starting with 12.7 ml of 4.10 × 10−2 M BaCl2, how many grams of ZnSO4 can be added to the solution?

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Starting with 12.7 ml of 4.10 × 10−2 M BaCl2, how many grams of ZnSO4 can be added to the solution?

before BaSO4 begins to precipitate ?

2007-03-23 08:20:08 · 1 answers · asked by Dd 1 in Science & Mathematics ➔ Chemistry

1 answers

BaCl2+SO4Zn ---> BaSO4+ZnCl2

Molecular mass Zn SO4 = 65.4 +32 +64 = 161.4

calculate number of moles BaCl2 in 12.7mL 4.1*10^-2

4.1 10^-2* 12.7*10^-3 =5.2*10^-4 moles

as you see on the equation 1 mole BaCl2 corresponds to one mole ZnSO4 . You nee 5.2 10^-4mole

so in grams 5.2*10^-4 *161.5=0.084g

2007-03-23 08:31:36 · answer #1 · answered by maussy 7 · 0⤊ 1⤋