What volume of sulfur dioxide, SO2 (at STP), is produced by the burning of 0.250 mole of sulfur?

S(s) + O2(g) → SO2(g)

5.6 L is the final volume for SO₂ at STP condtions

Explanation:

The reaction is:  S(s) + O₂ (g) →  SO₂ (g)

Ratio for this reaction is 1:1, if we see stoichiometry.

1 mol of sulfur can react to 1 mol of oxygen in orden to produce 1 mol of sulfur dioxide.

Then, 0.250 moles of sulfur can produce 0.250 moles of SO₂.

If we solve this by a rule of three, we can say:

At STP 1 mol occupies 22.4L

Then, 0.250 moles will be contained at (0.250 mol . 22.4L) /1mol = 5.6 L

If we apply the Ideal Gases Law:

P . V = n . R . T

1 atm . V = 0.250 mol . 0.082 . 273K

V = (0.250 mol . 0.082 . 273K) / 1atm → 5.6 L

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.