Question

The annual production of sulfur dioxide from burning coal and fossil fuels, auto exhaust, and other sources is about 26 million tons. The equation for the reaction is s(s) + o2(g) → so2(g) how much sulfur (in tons), present in the original materials, would result in that quantity of so2?

Answer 1

Answer is: quantity of sulfur is 13 tons.

Chemical reaction: S(s) + O₂(g) → SO₂(g).

From chemical reaction: n(S) : n(SO₂) = 1 : 1.

n(S) = n(SO₂); amount of substance.

m(S) ÷ M(S) = m(SO₂) : M(SO₂).

m(S) : 32 g/mol = 26 t : 64 g/mol.

m(S) = (32 g/mol · 26 t) ÷ 64 g/mol.

m(S) = 13 t = 13000 kg; mass of sulfur.

Answer 2

Answer: The mass of sulfur produced is
`1.305* 10^(-4)tons`

Step-by-step explanation:

We are given:

Mass of sulfur dioxide = 26 million tons =
`235.872* 10^(11)g` (Conversion factor:
`\text{1 million ton}=9.072* 10^(11)g` )

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Given mass of sulfur dioxide =
`235.872* 10^(11)g`

Molar mass of sulfur dioxide = 64 g/mol

Putting values in equation 1, we get:

`\text{Moles of sulfur dioxide}=(235.872* 10^(11)g)/(64g/mol)=3.70* 10^(11)mol`

For the given chemical equation:

`S(s)+O_2(g)\rightarrow SO_2(g)`

By Stoichiometry of the reaction:

1 mole of sulfur dioxide is produced by 1 mole of Sulfur.

So,
`3.70* 10^(11)` moles of sulfur dioxide will be produced by =
`(1)/(1)* 3.70* 10^(11)=3.70* 10^(11)` moles of sulfur.

Calculating the mass of sulfur by using equation 1, we get:

Molar mass of sulfur = 32 g/mol

Moles of sulfur =
`3.70* 10^(11)mol`

Putting values in equation 1, we get:

`3.70* 10^(11)mol=\frac{\text{Mass of sulfur}}{32g/mol}\\\\\text{Mass of sulfur}=(32g/mol* 3.70* 10^(11)mol)=118.4* 10^(11)g`

Converting this into tons, we use the conversion factor:

1 ton = 907185 grams

So,
`118.4* 10^(11)g* (1ton)/(907185g)=1.305* 10^(-4)tons`

Hence, the mass of sulfur produced is
`1.305* 10^(-4)tons`