Consider the reaction between NaCl and AgNO3 to form silver chloride. How much of the insoluble precipitate will be recovered if 582.4 g of AgNO3 reacts with excess NaCl?

The correct answer is: 491.2 g AgCl

Explanation:

The balanced chemical equation for the reaction between AgNO₃ and NaCl is the following:

AgNO₃(aq) + NaCl(aq) ⇒ AgCl(s) + NaNO₃(aq)

From the products, silver chloride (AgCl) is an insoluble salt, so it will precipitate as is formed. Moreover, from the balanced equation we can see that 1 mol AgCl is obtained from 1 mol AgNO₃:

1 mol AgCl = 107.8 g + 35.4 g = 143.2 g

1 mol AgNO₃= 107.8 g + 14 g + (3 x 16 g) = 169.8 g

Thus, we have a stoichiometric ratio of 143.2 g AgCl/169.8 g AgNO₃. From the reaction of 582.4 g AgNO₃, the following mass of AgCl will be formed:

582.4 g AgNO₃ x 143.2 g AgCl/169.8 g AgNO₃ = 491.2 g AgCl

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2