Final answer:
After the reaction, there will be 12.0 g of calcium fluoride, 6.76 g of dinitrogen monoxide, and 5.53 g of water.
Step-by-step explanation:
To calculate the mass of each substance after the reaction, we need to use stoichiometry. From the balanced chemical equation, we can see that 1 mole of calcium nitrate reacts with 2 moles of ammonium fluoride to form 1 mole of calcium fluoride, 1 mole of dinitrogen monoxide, and 2 moles of water.
1. Calculate the number of moles of calcium nitrate:
25.24 g / (164.10 g/mol) = 0.1535 mol
2. Calculate the number of moles of ammonium fluoride:
26.30 g / (37.04 g/mol) = 0.7096 mol
3. Determine the limiting reactant. Since 1 mole of calcium nitrate reacts with 2 moles of ammonium fluoride, we compare the number of moles of calcium nitrate to ammonium fluoride.
0.1535 mol of calcium nitrate * (2 mol ammonium fluoride / 1 mol calcium nitrate) = 0.3070 mol of ammonium fluoride
Since the number of moles of ammonium fluoride is greater than the number of moles of calcium nitrate, the ammonium fluoride is the limiting reactant.
4. Calculate the number of moles of each product:
Since 1 mole of ammonium fluoride reacts with 1 mole of calcium fluoride, dinitrogen monoxide, and 2 moles of water, we can use the ratio of the coefficients in the balanced chemical equation to calculate the number of moles of each product.
0.3070 mol of ammonium fluoride * (1 mol calcium fluoride / 2 mol ammonium fluoride) = 0.1535 mol of calcium fluoride
0.3070 mol of ammonium fluoride * (1 mol dinitrogen monoxide / 2 mol ammonium fluoride) = 0.1535 mol of dinitrogen monoxide
0.3070 mol of ammonium fluoride * (2 mol water / 2 mol ammonium fluoride) = 0.3070 mol of water
5. Calculate the mass of each product:
0.1535 mol of calcium fluoride * (78.08 g/mol) = 12.0 g of calcium fluoride
0.1535 mol of dinitrogen monoxide * (44.02 g/mol) = 6.76 g of dinitrogen monoxide
0.3070 mol of water * (18.02 g/mol) = 5.53 g of water