Suppose a 32.6 g sample of CaSiO3 is reacted with 30.1 L of HF at 27.0 °C and 1.00 atm. Assuming the reaction goes to completion, calculate the mass (in grams) of the SiF4 and H2O produced in the reaction.

The balanced chemical equation for the reaction can be written as:
CaSiO_3 + 6HF -> SiF_4 + 3H_2O + CaF_2
 
The molar masses of the species of interest are:
CaSiO3 = 116.16 g/mol
SiF4 = 104.08 g/mol
H2O = 18 g/mol
 
Using the given data, moles of CaSiO3 = 32.6 g/ 116.16 g/mol = 0.281 moles
 
Similarly, moles of HF = PV/RT (using the gas law) 
 = (1 atm x 30.1 l)/(0.0821 l atm/mol/K x (27 + 273) K) = 1.22 moles
 
Using stoichiometry, 1 moles of CaSiO3 reacts with 6 moles of HF.
i.e., 0.281 moles of CaSiO3 will react with 6 x 0.281 moles = 1.686 moles of HF.
Since the available amount of HF is less than 1.686 moles, HF is the limiting reactant. 
 
Again using stoichiometry, 6 moles of HF produces 1 moles of SiF4.
Hence, the moles of SiF4 produced = 1/6 x 1.22 moles = 0.203 moles
and the amount of SiF4 produced = 0.203 moles x 104.08 g/moles = 21.13 g.
 
Similarly, the amount of water produced = 3/6 x 1.22 moles x 18 g/mol = 10.98 g.
 
Hope this helps.