Consider the following reaction: Cr(NO3)3 (aq) + 2NaF (aq) --> 3NaNO3 (aq) + CrF3 (s)
If 21.0 grams of NaF are needed to precipitate all of the Cr+3 ions present in 0.125L of a solution of Cr(NO3)3, what is the molarity of the Cr(NO3)3 solution?

Your answer should be to 2 decimal places.

2.01

Explanation:

First, let's convert grams to moles

(Na) 22.99 + (F) 18.998 = 41.988

Every mole of NaF is  41.988 grams

21/41.988 = 0.500143 moles of NaF

For every Cr+3, we will need 2 NaF, so Cr+3 will be half of NaF

0.500143/2 = 0.250071

molarity = moles/liters

0.250071/0.125 = 2.0057 M

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 :

the temperature of the gas, in k, not celsius.