Homework Help: Chemistry

Posted by Jo on Monday, April 23, 2012 at 1:53pm.

A 12 mL sample of 0.4 M silver nitrate solution is mixed with 8 mL of 0.8 M sodium chloride solution. Determine the identity of precipitate and the number of grams produced.

Chemistry - DrBob222, Monday, April 23, 2012 at 2:14pm
12 mL x 0.4M AgNO3 = 4.8 millimols.
8 mL x 0.8M NaCl = 6.4 mmols.
AgNO3 + NaCl => AgCl + NaNO3
This is a limiting reagent problem since amounts are given for BOTH reactants. You can tell, since the reactants and products are 1:1:1:1 that AgNO3 is the limiting reagent so you will produce 4.8 mmillimols (0.0048 mols) AgCl and convert that to grams. g = mols x molar mass.

Answer this Question

Related Questions

Chemistry - Silver(I) nitrate and sodium sulfate react to form silver(I) sulfate...
chem. - Silver(I) nitrate and sodium sulfate react to form silver(I) sulfate and...
Chemistry - when 20 ml of a .10 M silver nitrate solution is mixed with 25 ml ...
Chemistry - it takes 37.50 mL of 0.152 M sodium chromate to titrate 25.00 mL of ...
Chem - Solutions of sodium carbonate and silver nitrate react to form solid ...
chemistry - What is the silver ion concentration in a solution prepared by ...
Chemistry - Assignment  Calculate the pH, pOH, [H+], and [OH&#...
Chemistry - What is the silver ion concentration in a solution prepared by ...
Chemistry - Please help! - What is the silver ion concentration in a solution ...
Chemistry - A reaction is peformed in which 50 ml of .400M silver nitrate and 75...

For Further Reading

First Name:
School Subject:
Answer: