how do I calculate the mass of MgCO3 precipitated by mixing 10.0 ml of a 0.2 M Na2CO3 solution with 5mL of a .03 M Mg(NO3)2 solution?
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This turns out to be a limiting reactant problem.
Na2CO3 +Mg(NO3)2 --------> MgCO3 + 2NaNO3
0.010L*0.2M Na2CO3=0.002m
0.005L*0.03M Mg(NO3)2 =0.00015m
The equation shows that the reactants react on a 1:1 ratio so the limiting reactant is the 0.00015m Mg(NO3)2
0.00015m Mg(NO3)2 =0.00015m MgCO3 * 84.31 =0.0126g MgCO3
Na2CO3 +Mg(NO3)2 --------> MgCO3 + 2NaNO3
0.010L*0.2M Na2CO3=0.002m
0.005L*0.03M Mg(NO3)2 =0.00015m
The equation shows that the reactants react on a 1:1 ratio so the limiting reactant is the 0.00015m Mg(NO3)2
0.00015m Mg(NO3)2 =0.00015m MgCO3 * 84.31 =0.0126g MgCO3