Question

Ca(OH)2(k) + NH4Cl(k) → CaCl2(aq) + NH3(g) + H2O(s) (Reaction is not balanced)

If a mixture containing 33 g NH4Cl and 33 g Ca(OH)2 is heated, how many grams of NH3 will form and which reactant remains in excess, and in what mass? (Ca: 40 g/mole, Cl: 35.5 g/mole, O: 16 g/mole, N: 14 g/mole, H: 1 g/mole)

Answer 1

Step-by-step explanation:

First, we need to balance the equation:

Ca(OH)2(s) + 2NH4Cl(s) → CaCl2(aq) + 2NH3(g) + 2H2O(l)

Now, we can use stoichiometry to determine which reactant is in excess and how much NH3 will form.

From the balanced equation, we can see that 1 mole of Ca(OH)2 reacts with 2 moles of NH4Cl to produce 2 moles of NH3. We need to find the limiting reactant in the given mixture of 33 g NH4Cl and 33 g Ca(OH)2.

The molar mass of NH4Cl is 53.5 g/mol (1N + 4H + 1Cl), so 33 g NH4Cl is equal to:

33 g / 53.5 g/mol = 0.617 moles NH4Cl

The molar mass of Ca(OH)2 is 74 g/mol (1Ca + 2O + 2H), so 33 g Ca(OH)2 is equal to:

33 g / 74 g/mol = 0.446 moles Ca(OH)2

From the balanced equation, we know that 1 mole of Ca(OH)2 reacts with 2 moles of NH4Cl, so the maximum number of moles of NH4Cl that can react with 0.446 moles of Ca(OH)2 is:

0.446 moles Ca(OH)2 x (2 moles NH4Cl / 1 mole Ca(OH)2) = 0.892 moles NH4Cl

Since we have only 0.617 moles of NH4Cl, NH4Cl is the limiting reactant and Ca(OH)2 is in excess.

To calculate the amount of NH3 produced, we can use the stoichiometric ratio from the balanced equation. For every 2 moles of NH3 produced, we need 2 moles of H2O. The molar mass of NH3 is 17 g/mol (1N + 3H), so 0.617 moles of NH3 is equal to:

0.617 moles x 2 moles NH3 / 2 moles H2O x 17 g/mol NH3 = 10.5 g NH3

Therefore, 10.5 g of NH3 will form, and Ca(OH)2 is in excess with a mass of:

33 g Ca(OH)2 - (0.446 moles Ca(OH)2 x 74 g/mol Ca(OH)2) = 0.26 g Ca(OH)2