A solution is created by dissolving 13.0 grams of ammonium chloride in enough water to make 325 mL of solution. How many moles of ammonium chloride are present in the resulting solution?

0.243 moles of ammonium chloride are present in the resulting solution

Explanation:

We determine our compounds of the solution:

Solute: NH₄Cl

The solvent is water

Solution's volume: 325 mL

If we want to know, how many moles of ammonium chloride are present in the resulting solution we may just convert the solute's mass to moles.

Molar mass NH₄Cl = 53.45 g/mol

Moles of NH₄Cl = Mass of NH₄Cl / Molar mass NH₄Cl

13 g / 53.45 g/mol = 0.243 moles

We have 0.243 moles of ammonium chloride is the solution

Explanation:

Step 1: Data given

Ammonium chloride = NH4Cl

Mass of ammonium chloride = 13.0 grams

Volume of water = 325 mL = 0.325 L

Molar mass of ammonium chloride

⇒ Molar mass N = 14.0 g/mol

⇒ Molar mass of H = 1.01 g/mol : 4*1.01 = 4.04 g/mol

⇒Molar mass of Cl = 35.45 g/mol

Molar mass NH4Cl = 53.49 g/mol

Step 2: Calculate moles ammonium chloride

Moles NH4Cl = mass NH4Cl / molar mass NH4Cl

Moles NH4Cl = 13.0 grams / 53.49 g/mol

Moles NH4Cl = 0.243 moles

We have 0.243 moles of ammonium chloride is the solution

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.