If 0.40 mol of h₂ and 0.15 mol of o₂ were to react as completely as possible to produce h₂o, what mass of reactant would remain?

0.1 mole of H2, that in grams will represent: 1 mol = 2 grams H2,

0.1 moles = 0.2 grams of H2

Explanation:

The complete reaction will it be:

2 H2 + O2  →  2  H2O

1.- That means that for 1 mole of O2 we produce 2 moles of H2O but we just have 0.15 moles of O2 so if we apply a rule of three:

moles of H2O to produce = (0.15  x 2 )/1 = 0.3 moles produced of H2O

2.- So as we can observe in the chemical reaction, you need 2 moles of H2 to produce 2 moles of H2O but you can just produce 0.3 moles, so again if you do a rule of three:

moles of H2 that are going to be used: (0.3 x 2) / 2 = 0.3 moles of H2

3.- Since you have 0.4 moles, but you are just going to use 0.3 , you left 0.1 moles

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 :