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Veronika [31]
3 years ago
6

When sulfur burns, it forms sulfur dioxide (SO2). Its chemical reaction is S + O2 → SO2.

Chemistry
1 answer:
Elenna [48]3 years ago
4 0

Answer:

The mass of SO2 will be equal to the sum of the mass of S and O2.

Explanation:

This can be explained by the <em>Law of Conservation of Mass</em>. This law states that mass can neither be created nor destroyed. Knowing this, we can say that the reactants of a chemical reaction must be equal to the products.

In this case, the reactants Sulfur (S) and Oxygen (O2) must equal the mass of the product Sulfur Dioxide (SO2). Therefore, the statement <em>"The mass of SO2 will be equal to the sum of the mass of S and O2" </em>is correct.

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The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t
slava [35]

<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 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)

Given mass of iron(III) phosphate = 20.00 g

Molar mass of iron(III) phosphate = 150.82 g/mol

Putting values in equation 1, we get:

\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol

The given chemical equation follows:

2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = \frac{1}{2}\times 0.133=0.0665moles of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

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3 years ago
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Degree of unsaturation = [ 12 - 6 ] / 2 = 3

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Using the formular above;

Degree of unsaturation = [ 2(10) + 2 - 6 -2 ] / 2

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Using the formular above;

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