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mihalych1998 [28]
3 years ago
15

Sodium peroxide (Na2O2) is used to remove carbon dioxide from (and add oxygen to) the air supply in spacecrafts. It works by rea

cting with CO2 in the air to produce sodium carbonate (Na2CO3) and O2. 2 Na2O2(s) + 2 CO2(g) → 2 Na2CO3(s) + O2(g) What volume (in liters) of CO2 can be consumed at STP by 435 g Na2O2?
Chemistry
1 answer:
bazaltina [42]3 years ago
6 0

 The volume  (in liters)  of CO₂  that can  be consumed at STP by 435 g Na₂O₂  is  125 L of Co₂

<u><em>calculation</em></u>

2Na₂O₂(s)  +2 CO₂ (g)→  2 Na₂CO₃(s)  + O₂(g)

Step 1 : find the moles  of Na₂O₂

moles = mass÷  molar mass

from periodic table the  molar mass  of Na₂O₂ = (23 x2) +( 16 x2) = 78 g/mol

 moles= 435 g÷ 78 g/mol = 5.58 moles

Step 2: use the mole ratio to determine the moles of CO₂

from given equation  Na₂O₂ : CO₂ =2 :2 =1:1

Therefore the  moles of CO₂  is also = 5.58 moles

Step 3: find the volume  of CO₂  at STP

that is at STP      1  mole of a gas = 22.4 L

                          5.58 moles = ? l

<em>by cross multiplication</em>

= (5.58 moles x 22.4 L) / 1 mole = 125 L

 

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What is the vapor pressure of the solution if 35.0 g of water is dissolved in 100.0 g of ethyl alcohol at 25 ∘C? The vapor press
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<u>Answer:</u> The vapor pressure of the solution is 43.55 mmHg

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For water:</u>

Given mass of water = 35.0 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

\text{Moles of water}=\frac{35.0g}{18g/mol}=1.944mol

  • <u>For ethyl alcohol:</u>

Given mass of ethyl alcohol = 100.0 g

Molar mass of ethyl alcohol = 46 g/mol

Putting values in equation 1, we get:

\text{Moles of ethyl alcohol}=\frac{100.0g}{46g/mol}=2.174mol

Total moles of solution = [1.944 = 2.174] moles = 4.118 moles

  • Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

<u>For water:</u>

\chi_{\text{water}}=\frac{n_{\text{water}}}{n_{\text{water}}+n_{\text{ethyl alcohol}}}

\chi_{water}=\frac{1.944}{4.118}=0.472

<u>For ethyl alcohol:</u>

\chi_{\text{ethyl alcohol}}=\frac{n_{\text{ethyl alcohol}}}{n_{\text{water}}+n_{\text{ethyl alcohol}}}

\chi_{\text{ethyl alcohol}}=\frac{2.174}{4.118}=0.528

Dalton's law of partial pressure states that the total pressure of the system is equal to the sum of partial pressure of each component present in it.

To calculate the vapor pressure of the solution, we use the law given by Dalton, which is:

P_T=\sum_{i=1}^n (p_i\times \chi_i)

Or,

P_T=[(p_{\text{water}}\times \chi_{\text{water}})+(p_{\text{ethyl alcohol}}\times \chi_{\text{ethyl alcohol}}

We are given:

Vapor pressure of water = 23.8 mmHg

Vapor pressure of ethyl alcohol = 61.2 mmHg

Putting values in above equation, we get:

p_T=[(23.8\times 0.472)+(61.2\times 0.528)]\\\\p_T=43.55mmHg

Hence, the vapor pressure of the solution is 43.55 mmHg

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