Can you please help me with this chemistry problem
Favorites|Homepage
Subscriptions | sitemap
HOME > > Can you please help me with this chemistry problem

Can you please help me with this chemistry problem

[From: ] [author: ] [Date: 12-03-07] [Hit: ]
00015m MgCO3 * 84.31 =0.......
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?

-
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
1
keywords: with,Can,this,you,problem,please,chemistry,me,help,Can you please help me with this chemistry problem
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .