a 6.0 mole sample of Ca3(PO4)2 and a 12.0 mole sample of SiO2 react in an evacuated vessel. after one of the reactants is fully consumed, how much of the gas is created?

How exactly would you solve this problem? Would you have to find the molar mass of both sides and divide by a number?

To solve this problem, we need to determine the limiting reactant, which is the reactant that is completely consumed and thus determines the amount of product formed.

1. Start by writing the balanced chemical equation for the reaction:
Ca3(PO4)2 + 3 SiO2 → 3 CaSiO3 + 2 P2O5

2. Convert the given number of moles of both reactants to moles of the limiting reactant. Since the stoichiometric ratio is 1:3 for Ca3(PO4)2 and SiO2, respectively, we need to compare the number of moles of SiO2 with the number of moles of Ca3(PO4)2.

- Moles of SiO2: 12.0 mol (given)
- Moles of Ca3(PO4)2: 6.0 mol (given)

Since the stoichiometric ratio is 1:3, we need to multiply the moles of Ca3(PO4)2 by 3:
Moles of Ca3(PO4)2 = 6.0 mol × 3 = 18.0 mol

3. Determine the limiting reactant:
Compare the moles of SiO2 (12.0 mol) with the moles of Ca3(PO4)2 (18.0 mol). The reactant with the fewer number of moles is the limiting reactant, which in this case is SiO2.

4. Calculate the moles of product formed using the stoichiometric ratio from the balanced equation:
Since the stoichiometric ratio between SiO2 and the product is 3:3, the number of moles of the product formed is the same as the moles of the limiting reactant.

Moles of product = Moles of SiO2 (limiting reactant) = 12.0 mol

So, after the reaction, 12.0 moles of the gas product will be created.