Answer:
The actual yield is 2.675 grams and the theoretical yield is 4.0543 grams.
Explanation:
Based on the given question the reaction will be,
SiO₂ (s) + 4HF ⇒ SiF₄ + 2H₂O (l)
Let w be the concentration of silicon tetrafluoride generated by reacting silicon dioxide with hydrogen fluoride. The mass of silicon dioxide given in the question is 2.339 grams. The molar mass of silicon dioxide is 60 grams per mole and the molar mass of silicon tetrafluoride is 104 grams per mole.
Therefore, it can be said that 60 grams of silicon dioxide is giving rise to 104 grams of silicon tetrafluoride. So, 2.339 grams of silicon dioxide will generate,
As w is the weight of silicon tetrafluoride produced, so with the help of cross-multiplication we get,
60 × w = 2.339 × 104
w = 2.339 × 104 / 60 = 4.054 grams is the theoretical yield. However, the actual yield is 2.675 grams. So, the percent yield will be,
Percent yield = actual yield / theoretical yield × 100
Percent yield = 2.675 grams / 4.054 grams × 100
Percent yield = 65.98 %
The theoretical yield would be 4.052 g while the actual yield is 2.675 g
The reaction between SIO2 with HF can be represented by the following equation:
[tex]SiO_2 + 4HF ---> 2H_2O + SiF_4[/tex]
The ratio of SiO2 reacted to SiF4 produced is 1:1.
Theoretically:
mole of 2.339 SiO2 = mass/molar mass
= 2.339/60.08
= 0.00389 moles
Thus, equivalent mole of SiF4 = 0.00389 moles
Mass of 0.00389 moles of SiF4 = mole x molar mass
= 0.00389 x 104.0791
= 4.052 g
Hence, the theoretical yield is 4.052 g while the actual yield is 2.675 g
More on reaction yields can be found here: https://brainly.com/question/7786567?referrer=searchResults
