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Suppose that 50.0 mL of 0.12 M AgNO3 is added to 50.0 mL of 0.048 M NaCl solution. The mass of AgCl that forms?
Suppose that 50.0 mL of 0.12 M AgNO3 is added to 50.0 mL of 0.048 M NaCl solution. The mass of AgCl that forms is . The final concentration of the sodium ion is M while the concentration of the nitrate ion is M and the chloride concentration is M. The percentage of silver ions that have precipitated is %.
2 Answers
- maussyLv 71 decade agoFavorite Answer
To make such problems calculate the number of moles in each solution
50mL=0.05L of 0.12M AgNO3 contains 0.05*0.12=0.006moles of AgNO3
50mL of 0.048M of NaCl contains 0.05*0.048=0.0024 moles of NaCl
reaction of AgNO3 with NaCl is AgNO3+NaCl--> AgCl (solid) + NaNo3
so you see 1 molecule of AgNO3 reacts with 1 molecule NaCL
NaCL is the liliting reactant, you form 0.0024moles of AgCl
Molar mass AgCl 107.9+35.5= 143.4g
answer 143.5*0.0024=0.3342g
- ?Lv 44 years ago
Molarity = moles/liter. If we would like .5L of .2M answer, we would want: .2 M = ? moles/ .5L Rearrange this equation to get .2*.5=moles, or .a million moles mandatory. The molar mass of NaCl is fifty 8.40 5 g/mol. We only multiply our moles mandatory via the molar mass to get 5.80 5 grams of NaCl
