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Whitepunk [10]
3 years ago
8

What mass of O2 was used up in the reaction with an excess of SO2 gas if 14.2 g of sulfur trioxide is formed?

Chemistry
1 answer:
egoroff_w [7]3 years ago
7 0

Answer:

2.84g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2SO2 + O2 —> 2SO3

Step 2:

Determination of the mass of O2 that reacted and the mass of SO3 produced from the balanced equation.

This is illustrated below:

Molar mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 1 x 32 = 32g

Molar mass of SO3 = 32 + (16x3) = 80g/mol

Mass of SO3 from the balanced equation = 2 x 80 = 160g

Summary:

From the balanced equation above,

32g of O2 reacted to produce 160g of SO3.

Step 3:

Determination of the mass of O2 needed to produce 14.2g of SO3.

This can be achieved as shown below:

From the balanced equation above,

32g of O2 reacted to produce 160g of SO3.

Therefore, Xg of O2 will react to produce 14.2g of SO3 i.e

Xg of O2 = (32 x 14.2)/160

Xg of O2 = 2.84g

Therefore, 2.84g of O2 is needed to produce 14.2g of SO3.

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Answer:

V ∝ n  

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Suppose that pressure and temperature are constant.

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This is an example of <em>Avogadro's Law</em>: the volume of a gas is directly proportional to the number of moles (particles).

V ∝ n

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Part APart complete A sample of sodium reacts completely with 0.568 kg of chlorine, forming 936 g of sodium chloride. What mass
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Answer: 368 grams of sodium reacted.

Explanation:

The balanced reaction is :

2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)

\text{Moles of chlorine}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of chlorine}=\frac{0.568\times 1000g}{71g/mol}=8moles

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According to stoichiometry :

2 moles of NaCl are formed from = 2 moles of Na

Thus 16 moles of NaCl are formed from=\frac{2}{2}\times 16=16moles  of Na

Mass of Na=moles\times {\text {Molar mass}}=16moles\times 23g/mol=368g

Thus 368 grams of sodium reacted.

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