Stoichiometry...Chemistry HELP!?

3) In the first step in the production of sulfuric acid, sulfur is burned in the air to produce sulfur dioxide. How many moles of oxygen are required to react with 160 g of sulfur.

4) How many moles of hydrofluoric acid may be produced by the addition of sufficient sulfuric acid to 7.81 g o solid calcium fluoride?

*Could I have the steps and explanation please. I really don't understand.

Thank you so much.

Answer:
The first step is to have a balanced chemical equation.

In your first problem, it is: S + O2 == SO2. The balancing is easy, everything is one.

Now convert the grams of sulfur to moles, by dividing by the molar mass of S. Next, multiply by the molar ratio of 1 / 1 to get the moles of O2. Finally, multiply these moles by the molar mass of O2.

Your second equations is CaF2 + H2SO4 == 2HF + CaSO4

Once again, start by converting the 7.81 g of CaF2 to moles. Multiply by the molar ratio of 2 HF / 1 CaF2. This gives you moles of HF. Finally, multiply by the molar mass of HF and you are done
Lancenigo di Villorba (TV), Italy

QUESTION 3)
Some amount of sulphur undergoes burning by reaction with air's oxygen, the reaction is sketched below

S8(s) + 8 O2(g) ---> 8 SO2(g)

so I determine the Mole's Number involved in this reaction

160 / 256 = 0.625 mol

and I calculate the Exact Stoichiometric Amount of Oxygen

0.625 * 8 = 5.0 mol

BECAUSE ONE MOLE OF SULPHUR REACTS SINKING EIGHT MOLES OF OXYGEN.

QUESTION 4)
Another stoichiometric involves the following reaction

CaF2(s) + H2SO4(l) ---> CaSO4(s) + 2 HF(g)

producting Hydrofluoric Acid as dense, poisonous and corrosive fogs. The complete dissolution of Fluorite's 7.81g corresponding to a determined Mole's Number

7.81 / 78 = 0.1 mol

which aids me in the calculation involved to stoichiometric amount of Hydrofluoric Acid

0.1 * 2 = 0.2 mol

BECAUSE ONE MOLE OF CALCIUM FLUORIDE REACTS FORMING TWO MOLES OF HYDROFLUORIC ACID.

I hope this helps you.

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