pantomath asked in Science & MathematicsChemistry · 2 months ago

# Suppose that 26.7 mL of 0.490 M NaCl is added to 26.7 mL of 0.225 M AgNO3.?

Suppose that 26.7 mL of 0.490 M NaCl is added to 26.7 mL of 0.225 M AgNO3.

a) How many moles of AgCl would precipitate?

b) What are the concentrations in M of Ag+, Cl-, NO3- and Na+ in the reaction mixture after the reaction occurs?

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Lv 7
2 months ago

Moles of Cl- = 0.490 moles / L *0.0267 L = 0.01308 moles = moles Na+

moles of Ag+ = 0.225 mole/L * 0.0267 L = 0.00601 moles = moles NO3-

Ag++Cl- = AgCl

according to the equation we have a 1:1 ratio of reactants

we also see that Ag+ is the limiting reactant

so we will get  0.00601 moles of AgCl

the Ag+ will all precipitate so very little left in solution

moles of Cl- remaining =  0.01308 moles - 0.00601 moles = 0.00708 moles

these are in 2* 0.0267 L = 0.0534 L

[Cl-] = 0.00708 moles / 0.0534 L = 0.132 M

[NO3] = 0.00601 moles /  0.0534 L = 0.112 M

[Cl-] = 0.01308 moles / 0.0534 L = 0.245 M