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Yakvenalex [24]
3 years ago
9

The combustion reaction of propane is described by the reaction. c3h8 5o2 → 4h2o 3co2. how many moles of o2 are required to gene

rate 3 moles of co2
Chemistry
2 answers:
dmitriy555 [2]3 years ago
7 0
C₃H₈   +   5 O₂   →   4 H₂O   +   3 CO₂

mole ratio based on balance equation of  O₂   :   CO₂ i  s    5  :  3
                      C₃H₈   +   5 O₂   →   4 H₂O   +   3 CO₂

∴ if moles of CO₂  =  3 moles
  
then moles of O₂  = (3 moles ÷ 3) × 5
                             
                             =  <span>5 moles   </span>
Natali5045456 [20]3 years ago
6 0

Answer:

5 moles of O_{2} are required to generate 3 moles of CO_{2}

Explanation:

According to given balanced equation, 5 molecules of O_{2} reacts with 1 molecules of C_{3}H_{8} to produce 4 molecules of H_{2}O and 3 molecules of CO_{2}.

So, 5 molecules of O_{2} generate 3 molecules of CO_{2} or (5\times 6.023\times 10^{23}) molecules of O_{2} generate (3\times 6.023\times 10^{23}) molecules of CO_{2}

As 1 mol = 6.023\times 10^{23} molecules therefore  5 moles of O_{2} are required to generate 3 moles of CO_{2}

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An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?
umka2103 [35]

<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

K_b = molal boiling point elevation constant = 0.51°C/m

m_{solute} = Given mass of solute (CsCl) = 8.00 g

M_{solute} = Molar mass of solute (CsCl) = 168.4  g/mol

W_{solvent} = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC

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