Balancing equation is helpful because they help is figure out how much material

Question

Balancing equation is helpful because they help is figure out how much material to use and much we should except in the end. Say we had the reaction between F_o(NO_3) and MgCl_2 to make FeCl_1 and Mg(NO_3) Given 2.0 g of Fe(NO_3) and 0.50 g of MgCl_2, which will be the limiting reactant, and how much FeCl_3 should we expect to make? How many mL of 0.20 M MgCl_2 will be needed to react completely with 25 mL of 0.10 M Fe(NO_3)_3? Finally, we learned that if you are going to talk about what is "really" going on in solution, you need to write "net ionic equations". Here are a few examples:

Explanation / Answer

1) Moles of Fe(NO3)3 rected=mass/molar mass=2.0g/241.880 g/mol=0.00827 moles

moles of Mgcl2 reacted=0.5g/95.21/mol=0.00525 moles

MgCl2 is less in molar amount so it is the limiting reagent

from the balanced equation ,looking at the coefficients ,you can figure out

3 moles of MgCl2(limiting reagent ) reacts to give 2 moles FeCl3

So,0.00525 moles of MgCl2 will produce=2/3*0.00525 moles of FeCl3=0.00350 moles

mass of FeCl3 produced=0.00325 moles*162.2 g/mol=0.568 grams

2)moles of Fe(NO3)3=25 ml*0.1 mol/L=0.025 L*0.1 mol/L=0.0025 moles

2 moles Fe(NO3)3 completely rects with 3 moles MgCl2

So 0.0025 moles Fe(NO3)3 will react with MgCl2 =3/2*0.0025=0.00375 moles

so volume of MgCl2 required=moles/molarity=0.00375 moles/0.2 mol/L=0.01875 L=0.01875*1000=18.75 ml

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