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Svetach [21]
3 years ago
12

A gas occupies a volume of 67.54 liters at 158°C and 4.87 atm pressure. Calculate the number of moles of this gas.

Chemistry
1 answer:
Katarina [22]3 years ago
6 0

Answer:

The number of moles of the gas is 9.295 moles or 9.30 moles

Explanation:

We use PV = nRT

Where P = 4.87 atm;

V = 67.54 L

R= 0.0821Latm/molK

T = 158 C = 158 +273 K = 431 K

the number of moles can be obtained by substituting the values in the respective columns and solve for n

n = PV / RT

n = 4.87 * 67.54 / 0.0821 * 431

n = 328.9198 / 35.3851

n = 9.295moles

The number of moles is approximately 9.30moles.

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When 2.56 g of a compound containing only carbon, hydrogen, and oxygen is burned completely, 4.89 g of CO2 and 3.00 g of h2o are
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3 years ago
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Question 22
ruslelena [56]

Answer:

O2, oxygen.

Explanation:

Hello.

In this case, for the undergoing chemical reaction, we need to compute the moles of CO2 yielded by 85 g of CH4 (molar mass = 16 g/mol) and by 320 g of O2 (molar mass 32 g/mol) via the following mole-mass relationships:

n_{CO_2}^{by\ CH_4}=85gCH_4*\frac{1molCH_4}{16gCH_4} *\frac{1molCO_2}{1molCH_4} =5.3molCO_2\\\\n_{CO_2}^{by\ O_2}=320gO_2*\frac{1molO_2}{32gO_2} *\frac{1molCO_2}{2molO_2} =5molCO_2

Considering the 1:2:1 among CH4, O2 and CO2. Therefore, since 320 g of O2 yield the smallest amount of CO2 we infer that the limiting reactant is O2.

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