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borishaifa [10]
3 years ago
15

Convert 5.28 x 1019 molecules of C6H1206 to grams.

Chemistry
1 answer:
nlexa [21]3 years ago
5 0

Answer:

m=0.0158g

Explanation:

Hello there!

In this case, it is possible to comprehend these mass-particles problems by means of the concept of mole, molar mass and the Avogadro's number because one mole of any substance has 6.022x10²³ particles and have a mass equal to the molar mass.

In such a way, for C₆H₁₂O₆, whose molar mass is about 180.16 g/mol, the referred mass would be:

m=5.28x10^{19}molecules*\frac{1mol}{6.022x10^{23}molecules}*\frac{180.16g}{1mol}\\\\m=0.0158g

Best regards!

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The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2.
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Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.

Solution:- The balanced equation for the combustion of acetylene is:

2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

35.0molC_2H_2(\frac{4molCO_2}{2molC_2H_2})

= 70molCO_2

The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.

From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

35.0molC_2H_2(\frac{5molO_2}{2molC_2H_2})

= 87.5molO_2

Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.

So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

84.0molO_2(\frac{4molO_2}{5molO_2})

= 67.2molCO_2

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