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anzhelika [568]
2 years ago
9

At STP, 2.24 L of carbon dioxide gas is collected. What is the number of molecules of gas present in this volume?

Chemistry
1 answer:
Doss [256]2 years ago
8 0

Considering the definition of STP conditions and Avogadro's number, the number of molecules of gas present in 2.24 L is 6.023×10²² molecules.

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<h3>Definition of Avogadro's number</h3>

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.

<h3>Amount of molecules of carbon dioxide gas</h3>

At STP, 2.24 L of carbon dioxide gas is collected. So, you can apply the following rule of three: if by definition of STP conditions 22.4 L are occupied by 1 mole of carbon dioxide gas, 2.24 L are occupied by how many moles of carbon dioxide gas?

amount of moles of carbon dioxide gas=\frac{2.24 Lx1 mole}{22.4 L}

amount of moles of carbon dioxide gas= 0.1 moles

Finally, 0.1 moles of carbon dioxide gas are collected.

Then you can apply the following rule of three, considering the Avogadro's number:  If 1 mole of carbon dioxide gas contains 6.023×10²³  molecules, 0.1 mole of carbon dioxide gas contains how many molecules?

amount of molecules of carbon dioxide gas= (6.023×10²³ molecules × 0.1 moles)÷1 mole

amount of molecules of carbon dioxide gas=6.023×10²² molecules

Finally, the number of molecules of gas present in 2.24 L is 6.023×10²² molecules.

Learn more about

STP conditions:

brainly.com/question/26364483

brainly.com/question/8846039

brainly.com/question/1186356

Avogadro's Number:

brainly.com/question/11907018

brainly.com/question/1445383

brainly.com/question/1528951

#SPJ1

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How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 1.38 g of fe2s3 if the percent yield
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<span>Answer: 100 ml
</span>

<span>Explanation:


1) Convert 1.38 g of Fe₂S₃ into number of moles, n


</span>i) Formula: n = mass in grass / molar mass
<span>
ii) molar mass of </span><span>Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol
</span>

iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of <span>Fe₂S₃
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<span>2) Use the percent yield to calculate the theoretical amount:
</span>

<span>65% = 0.65 = actual yield/ theoretical yield =>


</span>theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol <span>Fe₂S₃</span><span>

3) Chemical equation:
</span>

<span> 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)


4) Stoichiometrical mole ratios:
</span>

<span>3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl


5) Proportionality:


</span>2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃
<span>
=> x = 0.020 mol FeCl₃


6) convert 0.020 mol to volume
</span>

<span>i) Molarity formula: M = n / V
</span>

<span>ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml
</span>

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