Calculate the volume occupied by 16.3 moles of nitrogen gas at 273 K and 101.3kPa

1 answer:

8 0

Answer: V= 366.3 L

Explanation:

Using the ideal gas law; PV = nRT

P= pressure = 101 kPa , n = Number of moles = 16.3 moles

V = Volume = ? R= Gas constant = 8.31 kPa L/Mol K

T= Temperature = 273K

We make ''V'' the subject of the formular;

V = n R T/ P = 16.3mol x 8.31 kPa. L / mol. K x 273 K

-----------------------------------------------------

101 kPa

V = 366.3 L

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Murrr4er [49]

Answer:

50.8 g

Explanation:

Equation of reaction.

$CH_4 + 2O_2 \to CO_2 + 2H_2O$

From the given information, the number of moles of methane = mass/ molar mass

= 15.4 g / 16.04 g/mol

= 0.960 mol

number of moles of oxygen gas = 90.3 g / 32 g/ mol

= 2.82 mol

Since 1 mol of methane requires 2 moles of oxygen

Then 0.960 mol of methane will require = 0.960 mol × 2 = 1.92 mol of oxygen gas

Thus, methane serves as a limiting reagent.

2.82 mol oxygen gas will result in 2.82 moles of water

So, the theoretical yield of water = moles × molar mass

= 2.82 mol × 18.01528 g/mol

= 50.8 g

8 0

3 years ago

romanna [79]

Answer:

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