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3 years ago

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The elements lithium and oxygen react explosively to form lithium oxide,Li2O. how many moles of lithium oxide will form if 4.37

mol of lithium tract?
4Li(s) +O2(g) -> 2Li2O(s)

1 answer:

anygoal [31]3 years ago

5 0

From the equation,

4 mole of lithium produces 2 mole of lithium oxide

4.37 mol of lithium produces (4.37÷4)x2 mol of lithium oxide

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How many grams of O2 are present in 44.1 L of O2 at STP?

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Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that occupies.
T is its temperature.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
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Then, in this case:

P= 1 atm
V= 44.1 L
n= ?
R= 0.082 $\frac{atmL}{molK}$
T= 0°C =273 K

Replacing in the expression for the ideal gas law:

1 atm× 44.1 L= n× 0.082 $\frac{atmL}{molK}$ × 273 K

Solving:

$n=\frac{1 atm x44.1 L}{0.082\frac{atmL}{molK}x273K}$

n=1.97 moles

Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:

$1.97 molesx\frac{32 g}{1 mole}$ = 63.04 g ≈ <u><em>63 g</em></u>

Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

Learn more about the ideal gas law:

<u>brainly.com/question/4147359?referrer=searchResults</u>

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