Iodine is prepared both in the laboratory and commercially by adding cl2(g)cl2(g) to an aqueous solution containing sodium iodide. 2nai(aq)+cl2(g)⟶i2(s)+2nacl(aq) 2nai(aq)+cl2(g)⟶i2(s)+2nacl(aq) how many grams of sodium iodide, nai, nai, must be used to produce 89.1 g89.1 g of iodine, i2? i2? mass: g nai

Answer : The mass of sodium iodide used to produced must be, 105.22 grams.

Explanation : Given,

Mass of = 89.1 g

Molar mass of = 253.8 g/mole

Molar mass of = 149.89 g/mole

First we have to calculate the moles of .

Now we have to calculate the moles of .

The balanced chemical reaction is,

From the balanced chemical reaction, we conclude that

As, 1 mole of obtained from 2 moles of

So, 0.351 moles of obtained from moles of

Now we have to calculate the mass of .

Therefore, the mass of sodium iodide used to produced must be, 105.22 grams.

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 :