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kykrilka [37]
3 years ago
7

How many grams of sodium oxide can be produced when 55.3 g Na react with 64.3 g O2?. . Unbalanced equation: Na + O2 → Na2O. . Sh

ow your solution and explain how you derive the final answer
Chemistry
2 answers:
a_sh-v [17]3 years ago
5 0
We are given with the equation 2Na + O2 → 2Na2O where Na and O2 react to form sodium oxide. We are also given with two amounts 55.3 g Na with 64.3 g O2 .In this case, we will find the limiting reactant. we divide each with the molar mass and the stoich coeff. Na is 1.202, O2 is 2.01. Hence the limiting is Na. we base the calculations here.The amount is equal to 46.88 grams
GenaCL600 [577]3 years ago
3 0
 <span>this is a limiting reagent problem. 

first, balance the equation 
4Na+ O2 ---> 2Na2O 

use both the mass of Na and mass of O2 to figure out how much possible Na2O you could make. 
start with Na and go to grams of Na2O 

55.3 gNa x (1molNa/23.0gNa) x (2 molNa2O/4 molNa) x (62.0gNa2O/1molNa2O) = 75.5 gNa2O 

do the same with O2 

64.3 gO2 x (1 molO2/32.0gO2) x (2 molNa2O/1 mol O2) x (62.0gNa2O/1molNa2O) = 249.2 g Na2O 

now you must pick the least amount of Na2O for the one that you actually get in the reaction. This is because you have to have both reacts still present for a reaction to occur. So after the Na runs out when it makes 75.5 gNa2O with O2, the reaction stops. 

So, the mass of sodium oxide is 

75.5 g</span>
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Trimix 10/50 is a gas mixture that contians 10% oxygen and 50% helium, and the rest is nitrogen. If a tank of trimix 10/50 has a
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Answer: 1.61 x 10⁴ kPa

Dalton's law <u>states that the sum of the partial pressures of each gas equals the total pressure of the gas mixture.</u> According to this law,

Pi = xi P

where Pi is the partial pressure of the gas i, xi is the mole fraction of the gas i in the gas mixture and P is the total pressure.

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xi = \frac{n_{i} }{n_{t} }

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To calculate the total number of moles of the mixture and thus determine the molar fraction of helium, we consider 100 g and calculate the number of moles that represent 10 g of O₂ (n₁), 50 g of He (n₂) and 40 g of N₂ (n₃):

n₁ =  10 g x \frac{1 mol}{31.998 g} = 0.313 mol

n₂ =  50 g x \frac{1 mol}{8.005 g} = 6.246 mol

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Then the total number of moles (nt) will be:

nt = n₁ + n₂ + n₃ = 0.313 mol + 6.246 mol +1.428 mol

nt = 7,987 mol

Then, the mole fraction of helium (x₂) in the mixture will be,

x₂ =  \frac{6.246 mol}{7.987 mol} = 0.78

and the partial pressure of helium in the mixture, according to Dalton's law, will be:

P₂ = x₂ P = 0.78 x 2.07 x 10⁴ kPa

P₂= 1.61 x 10⁴ kPa

So, <u>the partial pressure of helium if a tank of trimix 10/50 has a total pressure of 2.07 x 104 kPa is  1.61 x 10⁴ kPa</u>

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