`500 mL` of `0.2 M NaCl` sol. Is added to `100 mL` of `0.5 M AgNO_(3)` solution resulting in the formation white precipitate of `AgCl`. How many moles and how many grams of `AgCl` are formed? Which is the limiting reagent?

`500 mL` of `0.2 M NaCl` sol. Is added to `100 mL` of `0.5 M AgNO_(3)`

1.	`500 mL` of `0.2 M NaCl` sol. Is added to `100 mL` of `0.5 M AgNO_(3)` solution resulting in the formation white precipitate of `AgCl`. How many moles and how many grams of `AgCl` are formed? Which is the limiting reagent?
Answer» `underset({:(500 xx 0.2),(= 100 mmol):})(NaCl) + underset({:(100 xx 0.5),(= 50 mmol):})(AgNO_(3)) rarr AgCl + NaNO_(3)` 50 mmol of `AgNO_(3)` will react with 50 mmol of `NaCl` and 50 mmol of `AgCl` will be fromed. `AgCL = 50 m "moles" = 50 xx 10^(-3) = 0.05 mol` Weight of `AgCl = 0.05 xx 143.5 = 7.175 g` Hence, `AgNO_(3)` is the limiting reagent.

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`500 mL` of `0.2 M NaCl` sol. Is added to `100 mL` of `0.5 M AgNO_(3)` solution resulting in the formation white precipitate of `AgCl`. How many moles and how many grams of `AgCl` are formed? Which is the limiting reagent?

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