Calculate the mass of water produced from the reaction of 24.0 g of H2 and 160.0 g of O2. What is the limiting reagent?

  is the limiting reagent, 180 g of water is produced

Explanation:

To calculate the moles :

According to stoichiometry :

1 mole of require 2 moles of

Thus 5 moles of will require=  of

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

As 1 mole of give = 2 moles of

Thus 5 moles of give =  of

Mass of

Thus 180 g of will be produced from the given masses of both reactants.

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 :

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2