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...

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...

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.