1. How many grams are in 1.4 moles of Mg(NO3)2?

Explanation:

molar mass of the compound = 24 + 2× 14+ 16×6

= 24 + 28 + 96

= 148g/mol

no of moles = given mass / molar mass of the compound

1.4 = given mass/ 148

1.4×148=given mass

207.2 g

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 :