Consider the reaction of SiF4 with H2O to form HF and SiO2. If 4.13 g SiF4 is reacted with excess H2O and 2.14 g of HF is ultimately isolated, what is the percent yield for the reaction?

Question

Consider the reaction of SiF₄ with H₂O to form HF and SiO₂. If 4.13 g SiF₄ is reacted with excess H₂O and 2.14 g of HF is ultimately isolated, what is the percent yield for the reaction?

Percent yield = _____%

Answer 1

SiF₄ + 2H₂O ==> 4HF + SiO₂ ... balanced equation

Since H2O is present in excess, there is no need to find the limiting reactant, as it will be SiF₄

mols SiF₄ = 4.13 g x 1 mol SiF₄ / 104.1 g = 0.03967 mols SiF₄

Theoretical yield of HF = 0.03967 mols SiF₄ x 4 mols HF / mol SiF₄ x 20.0 g HF / mol HF = 3.17 g HF

Percent yield = 2.14 g / 3.17 g (x100) = 67.4%

Answer 2

Hi Arciene! I am linking to an image where I've worked out this percent yield. Let me know if you have questions.

https://imgur.com/t4dlMD2