sited Use the References to access important values if needed for this question.

Question

sited Use the References to access important values if needed for this question. For the following reaction, 13.6 grams of silicon tetrafluoride are allowed to react with 3.40 grams of water silicon tetrafluoride(s) water(l) hydrofluoric acid(aq) silicon dioxide (s) What is the maximum amount of hydrofluoric acid that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group No more group attempts remain

Explanation / Answer

number of moles of SiF4 = mass of SiF4 / molar mass of SiF4
= 13.6 / 104.1
=0.131 mol

number of moles of H2O = mass of H2O / molar mass of H2O
= 3.40 / 18
=0.189 mol

reaction is:
Si F4 + 2H2O ----> 4HF + SiO2

clearly H2O is limiting reagent
moles of HF formed = 2*moles of water
= 2*0.189
=0.378 mol

mass of HF formed = moles of HF * molar mass of HF
= 0.378 mol * 20.01
= 7.56 g
Answer: maximum amount of hydrofluoric acid = 7.56 g

limiting reagent is H2O

moles of SiF4 reacted = moles of water / 2 = 0.189/2 = 0.0945 mol

moles of SiF4 unreacted = 0.131 mol - 0.0945 mol = 0.0365 mol

mass of SiF4 unreacted = moles * molar mass
= 0.0365 mol * 104.1
=3.8 g
Answer: excess reagent = 3.8 g

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