Calcium nitrate and ammoonium fluoride react to form calcium fluoride, dinitrogenmonoxide and water vapor. What mass of each substance is present after 28.42 g of calcium nitrate and 29.60 g of ammonium fluoride react completely?
You have a limiting reagent problem when both reagents are listed with grams or moles.
1. Write the equation and balance it.
2a. Calculate moles Ca(NO3)2. moles = g/molar mass.
2b. Calculate moles NH4F.
3a. Convert moles of one of the reactants to moles of the other.
3b. Determine the limiting reagent.
3c. ALL of the limiting reagent will be used and there will be zero left.
3d. Subtract moles of the other reactant from initial moles.
3e. Calculate grams of the other reactant remaining.
Post your work if you get stuck.
28.42g/164.1= .17318 mol Ca(NO3)2
29.6g/37.042=.79909 mol (NH4)F
(1mol Ca(NO3)2 /.17318) x (2mol (NH4)F/x)=.34636
(2 mol (NH4)F/.79909)x(1 mol Ca(NO3)2/x)=.399545
so (NH4)F (.34636) is limiting reagent.
now I'm stuck. Not sure what you mean about subtracting moles from initial moles etc.....thanks!
To find the mass of each substance present after the reaction, we need to use stoichiometry and balance the equation first.
The balanced chemical equation for the reaction between calcium nitrate (Ca(NO3)2) and ammonium fluoride (NH4F) is:
3Ca(NO3)2 + 20NH4F → 12HF + 12NH3 + 3N2O + 3CaF2 + 10H2O
From the balanced equation, we can see that the molar ratio between calcium nitrate and calcium fluoride is 3:3 (or 1:1) and the ratio between ammonium fluoride and calcium fluoride is 20:3.
Step 1: Calculate the moles of calcium nitrate (Ca(NO3)2) and ammonium fluoride (NH4F) using their molar masses.
Molar mass of Ca(NO3)2 = 40.08 g/mol (Ca: 40.08 g/mol, N: 14.01 g/mol, O: 16.00 g/mol)
Molar mass of NH4F = 37.04 g/mol (N: 14.01 g/mol, H: 1.01 g/mol, F: 18.99 g/mol)
Moles of Ca(NO3)2 = mass / molar mass = 28.42 g / 164.09 g/mol = 0.1733 mol
Moles of NH4F = mass / molar mass = 29.60 g / 37.04 g/mol = 0.7982 mol
Step 2: Use the ratio obtained from the balanced equation to determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of products formed.
According to the balanced equation, the ratio of Ca(NO3)2 to NH4F is 3:20.
So, for every 3 moles of Ca(NO3)2, we need 20 moles of NH4F.
Moles of NH4F / Ratio = 0.7982 mol / 20 = 0.03991 mol (actual moles of NH4F needed)
Since we only have 0.1733 mol of Ca(NO3)2, it is in excess, and NH4F is the limiting reactant.
Step 3: Calculate the moles of the other reactants consumed and products formed using the stoichiometry of the balanced equation.
Moles of Ca(NO3)2 consumed = 0.1733 mol
Moles of NH4F consumed = 0.03991 mol
Moles of CaF2 formed (using the ratio from the balanced equation) = 0.03991 mol × (3 mol CaF2 / 20 mol NH4F) = 0.005987 mol
Moles of N2O formed = 0.03991 mol × (3 mol N2O / 20 mol NH4F) = 0.005987 mol
Moles of H2O formed = 0.03991 mol × (10 mol H2O / 20 mol NH4F) = 0.01997 mol
Step 4: Calculate the mass of each substance using the moles and their molar masses.
Mass of CaF2 formed = moles × molar mass = 0.005987 mol × 78.08 g/mol = 0.468 g
Mass of N2O formed = moles × molar mass = 0.005987 mol × 44.01 g/mol = 0.2634 g
Mass of H2O formed = moles × molar mass = 0.01997 mol × 18.02 g/mol = 0.3598 g
Therefore, after the complete reaction of 28.42 g of calcium nitrate and 29.60 g of ammonium fluoride, we would have:
- 0.468 g of calcium fluoride (CaF2)
- 0.2634 g of dinitrogen monoxide (N2O)
- 0.3598 g of water vapor (H2O)