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3 years ago

How many moles are equal to 83.4 L of O2?

Chemistry

1 answer:

ladessa [460]3 years ago

8 0

Answer:

$3.72mol$

Explanation:

Hello,

In this case, we consider that at STP conditions (273 K and 1 atm) we know that the volume of 1 mole of a gas is 22.4 L, thereby, for 83.4 L, the resulting moles are:

$22.4L\rightarrow 1mol\\83.4L\rightarrow X\\X=\frac{83.4L*1mol}{22.4L}=3.72mol$

This is a case in which we apply the Avogadro's law which relates the volume and the moles as a directly proportional relationship.

Best regards.

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2 years ago

(a.) a 0.7549g sample of the compound burns in o2(g) to produce 1.9061g of co2(g) and 0.3370g of h2o(g).

Natali [406]

The individual mass of C, H and O in given sample are 0.5196 g, 0.0374 g and 0.1979 g respectively.

Moles of CO2 formed can be calculated as

= Mass of CO2 / Molar mass of CO2

= 1.9061 / 44 = 0.0433 moles

<h3>Calculation of no. of moles of carbon</h3>

Now, moles of C which is present in one mole of CO2 = 1 mole

Moles of C in 0.0433 moles of CO2 = 0.0433 moles

As we know that, molar mass of C = 12 g / mol

Mass of C in 0.7549 g of given sample can be calculated as

= 0.0433 × 12 =0.5196 g

Mass of H2O formed = 0.3370 g

Similarly, Molar Mass of H2O = 18 g / mol

Moles of H2O = 0.3370 / 18 = 0.0187 moles

Moles of H present in 1 mole of H2O = 2 moles

Moles of H present in 0.0187 mole of H2O = 2 × 0.0187 = 0.0374 moles

Molar mass of H = 1 g / mol

Mass of H contained in 0.7549 g of sample = 1 × 0.0374= 0.0374 g

Mass of O in 0.7549 g sample can be calculated as

= 0.7549 – [(Mass of C ) + (Mass of H) ]

= 0.7549 – [ (0.5196) + (0.0374) ]

= 0.1979 g

Thus, we calculated that the individual mass of C, H and O in given sample are 0.5196 g, 0.0374 g and 0.1979 g respectively.

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DISCLAIMER: THE above question is incomplete. Complete question is given below:

A 0.7549g sample of the compound burns in o2(g) to produce 1.9061g of co2(g) and 0.3370g of h2o(g). Calculate the individual mass of C, H and O in the given sample.

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