Hydrochloric acid can be prepared using the reaction described by the chemical equation: 2 NaCl(s) + H2SO4(l) > 2 HCl(g) + Na2 SO4(s). How many grams of HCl can be prepared from 393 g of H2SO4 and 4.00 moles of NaCl?

145.84 g

Explanation:

To answer this problem first we need to determine the limiting reactant, to do so we convert 393 g of H₂SO₄ into moles using its molar mass:

4.01 moles of H₂SO₄ would react completely with (4.01 * 2) 8.02 moles of NaCl. There are not as many NaCl moles, so NaCl is the limiting reactant.

Now we calculate how many HCl moles are formed, using the number of moles of the limiting reactant and the stoichiometric coefficients of the reaction:

Finally we convert HCl moles into grams, using its molar mass:

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 :