Question

106.5 g of HCl(g) react with an unknown amount of NH3(g) to produce 156.3 g of NH4Cl(s). How many grams of NH3(g) reacted? Is the law of conservation of mass observed in the reaction? Justify your answer.

Answer 1

In the given reaction, 49.8 grams of NH3 reacted. The law of conservation of mass is observed.

Step-by-step explanation:

To determine the amount of NH3(g) that reacted, we can start by finding the mole ratio between HCl and NH3 in the balanced chemical equation:

2NH3(g) + 2HCl(g) → NH4Cl(s)

From the balanced equation, we know that 2 moles of NH3 react with 2 moles of HCl to produce 1 mole of NH4Cl. Therefore, the mole ratio of NH3 to HCl is 1:1.

Using the given mass of HCl(g) and the molar mass of HCl, we can calculate the number of moles of HCl: 106.5 g HCl x (1 mol HCl / 36.46 g HCl) = 2.92 mol HCl

Since the mole ratio of NH3 to HCl is 1:1, the number of moles of NH3 that reacted is also 2.92 mol.

To calculate the mass of NH3 that reacted, we can use the molar mass of NH3: 2.92 mol NH3 x (17.03 g NH3 / 1 mol NH3) = 49.8 g NH3

Therefore, 49.8 grams of NH3 reacted. The law of conservation of mass is observed in the reaction because the total mass of the reactants (106.5 g HCl and the unknown mass of NH3) is equal to the mass of the product (156.3 g NH4Cl).