If 500.0 mL of 0.450 M sodium phosphate is reacted with an excess of iron (II) nitrate solution, how many grams of iron (II) pho
1 answer:
You might be interested in
a. 1,4332 g
b. 7.54~g
<h3>Further explanation</h3>Given
Reaction
MgCl2 (s) + 2 AgNO3 (aq) → Mg(NO3)2 (aq) + 2 AgCl (s)
20 cm of 2.5 mol/dm^3 of MgCl2
20 cm of 2.5 g/dm^3 of MgCl2
Required
the mass of silver chloride - AgCl
Solution
a. mol MgCl2 :
From equation, mol AgCl = 2 x mol MgCl2=2 x 0.05=0.1
mass AgCl(MW=143,32 g/mol)= 0.1 x 143,32=1,4332 g
b. mol MgCl2 (MW=95.211 /mol):
From equation, mol AgCl = 2 x mol MgCl2=2 x 0.0263=0.0526
mass AgCl(MW=143,32 g/mol)= 0.0526 x 143,32=7.54~g