50cm'3 of sulphur (iv) oxide were produced at s.t.p when some quantity of powdered sulphur were burnt in excess oxygen.state the equation for the reaction
Ans
S+O2g-2SO2
I would write it this way since you added the gaseous state to O2.
S(s) + O2(g) ==> SO2(g)
Also I don't know how your prof handles S; some show that S actually is S8 in the normal state. If that is the case then the equation would be
S8(s) + 8O2(g) --> 8SO2(g)
Your Answer Is Very Correct.
8so2<g>
Yes
Well, well, well... looks like we've got some sulfur getting all fired up! Here's the equation for the reaction you mentioned:
S + O2 → 2SO2
Yes, my friend, that's what happens when sulfur decides to have a hot date with oxygen. They get all cozy, and in the end, 50 cm³ of sulfur (IV) oxide are produced at standard temperature and pressure. Room for two, oxygen? Looks like the sulfur needed some extra space to show off its "SO2" dance moves!
To determine the equation for the reaction between powdered sulfur and excess oxygen to produce 50cm³ of sulfur (IV) oxide at standard temperature and pressure (STP), follow these steps:
Step 1: Write out the balanced equation for the reaction between sulfur and oxygen.
S + O₂ → SO₂
Step 2: Determine the stoichiometry of the reaction.
In the balanced equation, it shows that 1 mole of sulfur reacts with 1 mole of oxygen to produce 1 mole of sulfur dioxide (SO₂).
Step 3: Calculate the amount of substance of sulfur used.
The volume is given as 50 cm³, but we need to convert it to moles for our stoichiometry. To do this, you need to know the molar volume of an ideal gas at STP. The molar volume is approximately 22.4 L/mol.
50 cm³ = (50/1000) L = 0.05 L (convert cm³ to L)
0.05 L = 0.05/22.4 mol ≈ 0.00223 mol (using the molar volume at STP)
Step 4: Write the final balanced equation using the calculated amount of substance.
To find the equation, we multiply the balanced equation by the appropriate factor to match the produced amount of sulfur (IV) oxide.
The balanced equation shows that 1 mole of sulfur dioxide is produced when 1 mole of sulfur reacts with 1 mole of oxygen.
However, we have calculated that we have 0.00223 mol of sulfur.
Multiplying the original balanced equation by 0.00223, we get:
0.00223(S + O₂) → 0.00223(SO₂)
After simplifying, the equation becomes:
S + O₂ → 2SO₂
Hence, the balanced equation for the reaction is:
S + O₂ → 2SO₂