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kirill [66]
3 years ago
9

The reaction of sulfur with oxygen is shown below. 2s(s) + 3o2(g) → 2so3(g) calculate the mass of sulfur trioxide (in g) produce

d when 100.0 g of each reactant is present.
Chemistry
2 answers:
belka [17]3 years ago
7 0

<u>Answer:</u> The mass of sulfur trioxide produced will be 167 grams

<u>Explanation:</u>

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

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

  • <u>For sulfur:</u>

Given mass of sulfur = 100.0 g

Molar mass of sulfur = 32.07 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfur}=\frac{100.0g}{32.07g/mol}=3.12mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 100.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{100.0g}{32g/mol}=3.125mol

The given chemical equation follows:

2S(s)+3O_2(g)\rightarrow 2SO_3(g)

By Stoichiometry of the reaction:

3 moles of oxygen gas reacts with 2 moles of sulfur metal

So, 3.125 moles of oxygen gas will react with = \frac{2}{3}\times 3.125=2.08mol of sulfur metal

As, given amount of sulfur metal is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of oxygen gas produces 2 moles of sulfur trioxide.

So, 3.125 moles of oxygen gas will produce \frac{2}{3}\times 3.125=2.08mol of sulfur trioxide

Now, calculating the mass of sulfur trioxide from equation 1, we get:

Molar mass of sulfur trioxide = 80.1 g/mol

Moles of sulfur trioxide = 2.08 moles

Putting values in equation 1, we get:

2.08mol=\frac{\text{Mass of sulfur trioxide}}{80.1g/mol}\\\\\text{Mass of sulfur trioxide}=(2.08mol\times 80.1g/mol)=167g

Hence, the mass of sulfur trioxide produced will be 167 grams

Oksanka [162]3 years ago
3 0

The mass of sulfur trioxide (in g) produced when 100 g of each reactant is present is 166.64  grams

calculation

find the moles of each reactant

moles = mass/molar mass

moles of SO2= 100 g/ 64 = 1.5625 moles

moles of O2= 100/32= 3.125 moles

oxygen is the limiting reagent therefore by use of mole ratio of O2:SO3 which is  3:2 the moles of SO3=  3.125 x2/3=2.083 moles


mass of SO3=  molar mass x moles

            = 2.083 moles x80 g/mol= 166.64 grams


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