To find the mass of calcium hydroxide that will react with 21.4g of ammonium chloride, we need to use the stoichiometry of the balanced equation.
From the balanced equation, we can see that the coefficient of ammonium chloride (NH4Cl) is 2, and the coefficient of calcium hydroxide (Ca(OH)2) is 1. This means that 2 moles of NH4Cl react with 1 mole of Ca(OH)2.
First, let's convert the mass of ammonium chloride (NH4Cl) to moles:
Mass of NH4Cl = 21.4g
Molar mass of NH4Cl = 14.01g/mol (N) + 4(1.01g/mol)(H) + 1(35.45g/mol) (Cl)
= 53.49g/mol
Number of moles of NH4Cl = mass / molar mass
= 21.4g / 53.49g/mol
≈ 0.3995 mol
Since the ratio of NH4Cl to Ca(OH)2 is 2:1, the number of moles of Ca(OH)2 needed to react with 0.3995 mol of NH4Cl is also 0.3995 mol.
Next, let's find the molar mass of Ca(OH)2:
Molar mass of Ca(OH)2 = 40.08g/mol (Ca) + 2(1.01g/mol)(H) + 16.00g/mol (O)
= 74.10g/mol
Finally, let's calculate the mass of calcium hydroxide (Ca(OH)2) needed to react with 0.3995 mol:
Mass of Ca(OH)2 = number of moles × molar mass
= 0.3995 mol × 74.10g/mol
≈ 29.65g
Therefore, the mass of calcium hydroxide that will react with 21.4g of ammonium chloride is approximately 29.65g.
What is the mass of calcium hydroxide that will react with 21.4g of ammonium chloride according to the following equation
2NH4 Cl + Ca(OH)2—> 2NH3+CaCl2 +2H2O
1 answer