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

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :