Question

So let’s say you have a beaker that contains 78.8g of ammonium carbonate and to this you add 175 mL of 5.80 M hydrochloric acid according to the following chemical equation:

2HCl(aq) + (NH4)2CO3(s) -- > 2NH4Cl(aq) + H2O(l) + CO2(g) .DeltaH = -145 kJ

A) Draw the beaker at the end of the reaction - what would e in the beaker and/or around it. What would the products look like on a molecular/atom/ion level?

B) Why would water vapor pressure be involved in this reaction?

C)At STP, how many L of the gas is produced? (Ignoring water vapor)

D) Assuming the volume given in the problem remains the same, what is the molarity of the ionic compound in the reaction at the end of the reaction?

E) How many moles of excess reactant are left over at the end of the reaction?

F) How many L of your gas would be produced if the reaction was done at 25C at 0.992atm?

Answer 1

This is an experimental question.

A) At the end of the reaction three products are formed which is also mention in above given reaction that is ammonium chloride (aq), water(l), and carbondioxide (g).

At the end of the reaction the aqueous solution of ammonium chloride will remain in the beaker means the ammonium chloride salt and water will be present in beaker and carbon dioxide gas will evolved with effervescence.

B) Water vapour is involved in this reaction because one of the gaseous product is obtained in this reaction.

C) STP stands Standard temperature pressure means any value which is measured at 1atm pressure and 273.15K temperature.

At STP, the volume of gas produced will be 22.4L in 1 mol.

D) At the end of the reaction the ionic compound obtained is ammonium chloride which is dissociated into NH₄⁺ and Cl^- .

Molarity = no. of moles of product / volume of solution in liters

2 moles of NH4Cl is produced at STP. Then,

Molarity = 2/22.4

Molarity = 0.089 mol/litre.

E) no.of moles of HCl = molarity * volume taken in liters

Given, molarity = 5.80 and volume = 175ml = 0.175L

no of moles of HCl = 5.80 * 0.175 = 1.015 moles.

no. of moles = mass / molar mass (for HCl)

Mass of HCl taken = no. of moles * molar mass

Mass of HCl taken = 1.015 * 36.5 = 37.04g.

The mass taken of ammonium carbonate = 78.8g

In the reaction between HCl and ammonium carbonate, the ammonium carbonate is excess reactant because

2(HCl) = 37.04*2 = 74.095 will react with only 74g of ammonium carbonate.

Left ammonium carbonate is 78.8-74.095 = 4.705g.