Where are you stuck on this problem?
mols CaSiO3 = grams/molar mass = ?
Use PV = nRT to find mols HF = ?
Find limiting reagent.
CaSiO3(s) +6 HF (g)-> CaF2(aq) + SiF4 (g) + 3 H2O(l)
If a 65.6 g sample of CaSiO3 is reacted in an inert 83 L container at 25.5°C with HF at 1.00 atm pressure find the following.
a. the mass of water, and CaF2 produced in the reaction and the mass of HF that remains.
b. The partial pressure of SiF4 in the container after the temperature has returned to 25.5°C
2 answers
I found 3.39 mols of HF and 0.56 mols of CaSiO3 but I'm confused what is the limiting reactant. It would be CaSio3 because it has less moles than HF?