What volume of a 0.750 m solution of hydrochloric acid a solution of hcl can be prepared from the hcl produced by the reaction of 25.0 of nacl with an excess of sulphuric acid?
nacl + h2so4 ? hcl + nahso4

0.570 L

Explanation:

Given:

Concentration of the HCl solution = 0.750 M

Mass of NaCl in the reaction = 25.0

Number of moles of NaCl in the reaction = Mass of NaCl / Molar mass of NaCl

also,

Molar mass of NaCl = 22.98 + 35.45 = 58.43 grams

thus,

Number of moles of NaCl in the reaction = 25 / 58.43 = 0.4278 moles

now,

1 mol of NaCl reacts to form 1 mol of HCl

thus,

0.4278 moles of NaCl reacts to form 0.4278 mol of HCl

also,

Concentration = Moles / volume of solution

0.750 = 0.4278 / volume of solution

or

volume of solution = 0.570 L

20 drops hope this

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 :