Answer to Question #107995 in General Chemistry for Galilea Gonzalez

Question

2a). Barium nitrate is mixed with copper II sulfate. Write a balanced chemical reaction for this double replacement reaction. Consult solubility table to determine the precipitate.



2b) write the net ionic equation




2c). If 250mL of 0.205 M Ba(Na3)2 is mixed with 25.0 mL of 0.228 M CuSO2, what mass of barium sulfate May be produced ? What is the limiting reactant ?



Show all steps for each questions

Answer 1

$$Ba(NO_3)_{2{(aq)}} + CuSO_{4(aq)} = BaSO_{4(s)} + Cu(NO_3)_{2(aq)}$$

$$Ba^{+2}_{(aq)} + SO_{4(aq)}^{-2} = BaSO_{4(s)}$$

$$n(Ba(NO_3)) = cV = 0.205M*0.250L = 0.05125mol$$

$$n(CuSO_4)= cV = 0.228M*0.025L = 0.0057mol$$

Barium nitrate in excess, calculation for copper (II) sulfate:

$$m(BaSO_4) = nM = 0.0057mol*233.38g/mol = 1.33g$$