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balu736 [363]
3 years ago
13

Using the following equation for the combustion of octane, calculate the heat associated with the formation of 100.0 g of carbon

dioxide. The molar mass of octane is 114.33 g/mol. The molar mass of carbon dioxide is 44.0095 g/mol.
Chemistry
1 answer:
givi [52]3 years ago
7 0

Answer: Heat associated with the formation of 100.0 g of carbon dioxide is 1563.2 kJ.

Explanation:

Reaction equation will be as follows.

    2C_{8}H_{18} + 25O_{2} \rightarrow 16CO_{2} + 18H_{2}O;  \Delta H^{o}_{rxn} = -11018 kJ

Mass of CO_{2} = 100 g

Hence, moles of CO_{2} present will be calculated as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{100 g}{44.0095 g/mol}

                             = 2.27 mol

Therefore, heat produced by 2.27 mol for the given reaction will be calculated as follows.

         2.27 mol \times \frac{11018 kJ}{16 mol CO_{2}}

             = 1563.2 kJ

Thus, we can conclude that heat associated with the formation of 100.0 g of carbon dioxide is 1563.2 kJ.

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You take three compounds consisting of two elements and decompose them. To determine the relative masses of X, Y, and Z, you col
miss Akunina [59]

Answer:

a) LAW OF MULTIPLE PROPORTIONS

b) 0.095g, 0.71g, 0.285g respectively

c) X2Y, YZ15, X6Y

d) hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

Explanation:

a) The assumptions made in solving this questions is the application of the LAW OF MULTIPLE PROPORTIONS. The Law of multiple proportions states that if two elements A and B combine together to form more than one compound, then the several masses of A which chemically combine with a fixed mass of B is in a simple ratio.

for example, copper forms two oxides ; copper(I) oxide (CuO) and copper(ii) oxide(Cu2O), it is possible for the two samples of the oxides to be reduced to Cu by reacting with Hydrogen gas. as such, certain masses of oxygen combine separately with a fixed mass of Cu. then the ratios of Cu are then determined.

b) To calculate the relative masses, we take note of the three compounds given, they all have some amount of Y in them, hence we can use Y  as our relative mass, this implies that the relative mass of Y = 1g

mass of X = 0.4g

mass of Y = 4.2g

amount of X in 1g of Y = 0.4 x 1 /4.2

= 0.095g

for compound 2;

mass of Y = 1.4g

mass of Z = 1.0g

amount of Z in 1g of Y =1.0 x 1 /1.4

= 0.71g

for compound 3;

mass of X = 2.0g

mass of Y = 7.0g

amount of X in 1g of Y = 1 x 2/7

= 0.285g

c) Applying the law of multiple proportions; since elements X and Z combine with a fixed mass of Y, they must bear a simple ratio;

compound 1/compound 3 = 0.095/0.285

= 1/3

compound 1/compound 2 = 0.095/0.71

= 2/15

compound 2/ compound 3 = 0.71/0.285

= 5/2

formular for compound 1 = X2Y

formula for compound 2 = YZ15

formular for compound 3 = X6Y

d) from the formular X2Y, we can get the amount of each product in XY using the ratios

%of compound XY in X = mass of compound X / total Mass

= 0.2/4.4 = 4.5%

as such in a 21g of compound XY, %of compound Y = 1 - %of compound X = 95.5%

hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

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