One step in the industrial production of nitric acid is the reaction of ammonia with oxygen gas. 4NH3(g)+5O2(g)-->4NO(g)+6H2O(g) What is the volume of nitrogen monoxide gas that can be produced from 8L of oxygen gas.

Moles of nitrogen monoxide gas produced from 8.0 L of oxygen gas = 6.4 L of nitrogen monoxide gas

Explanation:

Equation of reaction:  4NH₃(g) + 5O₂(g) --> 4NO(g) + 6H₂O(g)

From the equation of reaction, 5 moles of oxygen gas produces 4 moles of nitrogen monoxide gas

1 mole of a gas at STP occupies a volume of 22.4 L

5 moles of oxygen gas occupies a volume of 5 * 22.4 L = 112 L

4 moles of nitrogen monoxide as occupies a volume of 4 * 22.4 L = 89.6 L

Therefore, 89.6 L of nitrogen monoxide gas are produced from 112.0 L of oxygen

Moles of nitrogen monoxide gas produced from 8.0 L of oxygen gas = 8.0 * (89.6/112

Moles of nitrogen monoxide gas produced from 8.0 L of oxygen gas = 6.4 L of nitrogen monoxide gas

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.