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Pie
3 years ago
6

Given below is the balanced equation for the combustion of decane. 2C10H22 (l) + 31O2 (g) --> 20CO2 (g) + 22H2O (g) What is t

he ratio of carbon dioxide molecules to oxygen molecules?
Chemistry
1 answer:
Aleks [24]3 years ago
3 0

Answer: The ratio of carbon dioxide molecules to oxygen molecules is 20 :31

Explanation:

Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The balanced combustion reaction is,:

2C_{10}H_{22}(l)+31O_2(g)\rightarrow 20CO_2(g)+22H_2O(g)

Thus the ratio of carbon dioxide molecules to oxygen molecules is 20 : 31

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The normal freezing point of a certain liquid
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\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}\times 1000}{\text{Molar mass of urea}\times \text{Mass of X liquid}}

where,

\Delta T_f = change in freezing point

\Delta T_s = freezing point of solution = -0.5^oC

\Delta T^o = freezing point of liquid X= 0.4^oC

i = Van't Hoff factor = 1  (for non-electrolyte)

K_f = molal freezing point depression constant of X = ?

m = molality

Now put all the given values in this formula, we get

[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}

k_f=4.12^oC/m

Therefore, the molal freezing point depression constant of X is 4.12^oC/m

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