Question

When silver nitrate reacts with barium chloride, silver chloride and barium nitrate are formed. How many grams of silver chloride are formed when 10.4 g of silver nitrate reacts with 15.0 g of barium chloride

Answer 1

8.771 grams of AgCl2 is formed from the reaction of 10.4 g of silver nitrate reacts with 15.0 g of barium chloride.

Explanation:

The balanced equation for the reaction between silver nitrate and barium chloride is shown as:

2AgNO3 + BaCl2⇒ 2AgCl

From the reaction and the quantity of reactants given the limiting reagent could be calculated.

For limiting reagent we calculate the amount of AgCl formed from the quantity of reactants.

15 gram of BaCl2 gives

number of moles= $\frac{weight }{atomic weight of one mole of the substance}$

number of moles = $\frac{15}{288.3}$

                             = 0.052 moles of BaCl2 is used

1 mole of BaCl2 yielded 2 moles of AgCl

0.052 moles of BaCl2 will yield x moles

$\frac{2}{1}$ = $\frac{x}{0.052}$

 = 0.104 moles of AgCl is formed.

10.4 grams of siver nitrate gives

number of moles= $\frac{weight }{atomic weight of one mole of the substance}$

number of moles = $\frac{10}{169.87}$

                            = 0.0612 moles of AgNO3

so, 2 moles of AgNO3 forms 2 mole of AgCl

       0.0612 moles will form x moles of AgCl

0.0612 moles of Agcl will be formed

It is concluded that the lowest number of moles of AgCl is produced was from AgNO3 so AgNO3 is a limiting reagent in the reaction.

Thus, the  least moles of AgCl formed will be converted to grams

0.0612 x 143.32

= 8.771 grams