[Shea]

How many g of silver chloride will be produced by reacting 10g of silver nitrate with sodium chloride?

AgNO3 + NaCl -> AgCl + NaNO3

1mol AgNO3 / 1mol AgCl 

170 g of AgNO3 = 1 mole of AgNO3

10 g of AgNO3  = 10 / 170 mole of AgNO3 = 0.059 mole of AgNO3

1 mole of AgCl = 143.5 g of AgCl

0.059 mole of AgCl  =143.5 * 0.059 g of AgCl = 8.467 g of AgCl

[Borek]

OK

[swati]

    You asked exactly the same question earlier under the topic Balancing
http://www.chemicalforums.com/index.php?topic=9851.0

Could someone teach me how to write the balanced equation for this question?

How many g of silver chloride will be produced by reacting 10g of silver nitrate with sodium chloride?

Molar weight of AgNO₃ = 170 g

170 g of AgNO₃ = 1 mole of AgNO₃
 10 g of AgNO₃  = 10 / 170 mole of AgNO₃
                            = 0.059 mole of AgNO₃

So , 0.059  mole of AgNO₃ are present .

0.059 mole of AgNO₃  -------->  0.059 mole of AgCl
1        mole of AgCl = 143.5 g of AgCl
0.059 mole of AgCl  =143.5 * 0.059 g of AgCl = 8.467 g of Ag Cl

In short
Number of moles = Mass of the compound / Molar Mass

Mass of the compound = Number of moles * Molar Mass

[Shea]

The teacher said it was wrong.

[swati]

The teacher said it was wrong.

But I think it is correct .

  Did he gave you the correct answer ?
If yes, then post it and if no then please ask him .

[Borek]

But I think it is correct .

Result is correct, perhaps it was a problem with significant digits (8.5g) or something.