Answer:
Calcium fluoride (CaF2): 11.06 g
Dinitrogen monoxide (N2O): 2.08 g
Hydrogen fluoride (HF): 5.68 g
Water vapor (H2O): 8.52 g
Step-by-step explanation:
We can start by writing the balanced chemical equation for the reaction:
3 Ca(NO3)2 + 10 NH4F → 6 HF + 3 CaF2 + N2O + 10 H2O
This equation tells us that 3 moles of calcium nitrate (Ca(NO3)2) react with 10 moles of ammonium fluoride (NH4F) to produce 3 moles of calcium fluoride (CaF2), 1 mole of dinitrogen monoxide (N2O), 6 moles of hydrogen fluoride (HF), and 10 moles of water (H2O).
We can use the molar masses of each substance to convert the given masses into moles, and then use the mole ratios from the balanced equation to determine the moles of each product formed. Finally, we can convert the moles of each product back into masses using their respective molar masses.
Molar masses:
Ca(NO3)2 = 164.1 g/mol
NH4F = 37.04 g/mol
CaF2 = 78.08 g/mol
N2O = 44.01 g/mol
HF = 20.01 g/mol
H2O = 18.02 g/molMasses given:
m(Ca(NO3)2) = 16.8 g
m(NH4F) = 17.50 g
Converting masses to moles:
n(Ca(NO3)2) = m(Ca(NO3)2) / M(Ca(NO3)2) = 16.8 g / 164.1 g/mol = 0.1022 mol
n(NH4F) = m(NH4F) / M(NH4F) = 17.50 g / 37.04 g/mol = 0.4729 mol
Using the mole ratios from the balanced equation, we can determine the number of moles of each product formed:
n(CaF2) = 3/10 * n(NH4F) = 0.1419 mol
n(N2O) = 1/10 * n(NH4F) = 0.0473 mol
n(HF) = 6/10 * n(NH4F) = 0.2837 mol
n(H2O) = 10/10 * n(NH4F) = 0.4729 mol
Finally, we can convert the moles of each product back into masses:
m(CaF2) = n(CaF2) * M(CaF2) = 0.1419 mol * 78.08 g/mol = 11.06 g
m(N2O) = n(N2O) * M(N2O) = 0.0473 mol * 44.01 g/mol = 2.08 g
m(HF) = n(HF) * M(HF) = 0.2837 mol * 20.01 g/mol = 5.68 g
m(H2O) = n(H2O) * M(H2O) = 0.4729 mol * 18.02 g/mol = 8.52 g
Therefore, the mass of each substance present after the reaction is:
Calcium fluoride (CaF2): 11.06 g
Dinitrogen monoxide (N2O): 2.08 g
Hydrogen fluoride (HF): 5.68 g
Water vapor (H2O): 8.52 g