For the following reaction, 8.43 grams of silicon tetrafluoride are allowed to react with 7.68 grams of water
Sif4 + 2H2O -> 4HF + SiO2
What is the maximum amount of hydrofluoric acid that can be formed? What is the formula for the limiting reactant? What amount of excess reactant remains after the reaction is complete?
First, we need to find the number of moles for both reactants:
Silicon tetrafluoride (SiF4): m = 8.43 grams
Molar mass: 1*28.09 (Si) + 4*19.00 (F) = 28.09 + 76.00 = 104.09 g/mol
Moles = mass/molar mass = 8.43/104.09 = 0.0810 moles
Water (H2O): m = 7.68 grams
Molar mass: 2*1.01 (H) + 16.00 (O) = 18.02 g/mol
Moles = mass/molar mass = 7.68/18.02 = 0.426 moles
Now we should check the mole ratio of SiF4 to H2O. According to the balanced equation, the mole ratio is 1:2.
Divide the moles of each reactant by their coefficients in the balanced equation:
For SiF4: (0.0810 moles) / 1 = 0.0810
For H2O: (0.426 moles) / 2 = 0.213
SiF4 has the lowest mole ratio, so it's the limiting reactant.
Now, we can find the maximum amount of hydrofluoric acid (HF) produced using the mole ratio:
Mole ratio of HF to SiF4 is 4:1, so we multiply the moles of SiF4 by 4.
Moles of HF = 0.0810 * 4 = 0.324 moles
To find the mass of HF produced, we use the molar mass of HF (20.01 g/mol):
Mass of HF = moles * molar mass = 0.324 * 20.01 = 6.48 g
Now we can find the remaining amount of the excess reactant (H2O):
We need 0.0810 * 2 = 0.162 moles of H2O for the reaction. Initially, we have 0.426 moles of H2O.
Remaining moles of H2O = 0.426 - 0.162 = 0.264 moles
To find the mass of the remaining excess reactant:
Mass of H2O = moles * molar mass = 0.264 * 18.02 = 4.75 g
In summary:
The maximum amount of hydrofluoric acid (HF) formed is 6.48 grams.
The limiting reactant is silicon tetrafluoride (SiF4).
The amount of excess reactant (water, H2O) remaining after the reaction is 4.75 grams.