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3 years ago

Calculate the amount of CO2 produced when 1.00 mole of propane (C3H8) is burned in the presence of 1.00 mole of oxygen.

Chemistry

1 answer:

Anna [14]3 years ago

4 0

Answer:

The balanced Chemical equation is

$1C_3 H_8 +5O_2 >3CO_2+4H_2 O$

Since quantity of two reactants are given Hence we find the moles of product using both.

The quantity of product produced is the one we consider from the limiting reactant.

Limiting reactant is the reactant which runs out first.

So, least product is produced from the limiting reactant.

Mole ratio of $C_3 H_8:CO_2$ is 1: 3

$1.00 \mathrm{mol} \mathrm{C}_{3} \mathrm{H}_{\mathrm{8}} \times \frac{3 \mathrm{mol} \mathrm{CO}_{2}}{1 \mathrm{mol} \mathrm{C}_{3} \mathrm{H}_{8}}=3 \mathrm{molCO}_{2}$

Mole ratio of $O_2:CO_2$ is 5: 3

$1.00 \mathrm{mol} \mathrm{O}_{2} \times \frac{3 \mathrm{mol} \mathrm{co}_{2}}{5 \mathrm{mol} \mathrm{O}_{2}}=0.6 \mathrm{mol} \mathrm{CO}_{2}$

(Least produced)

So, the limiting reactant is $O_2$

The amount of $CO_2$ formed is 0.600mol (Answer)

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One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

True or False

alexdok [17]

1. True 2. False 3.false 4. False 5. True

5 0

4 years ago

All the elements of a family in the periodic table have what feature in common?

SCORPION-xisa [38]

They all have the same number of electrons in the electron cloud.

4 0

3 years ago

Read 2 more answers

RATE LAW QUESTION !

vivado [14]

In general, we have this rate law express.:

$\mathrm{Rate} = k \cdot [A]^x [B]^y$
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation $\Delta \mathrm{Rate} = \Delta [B]^y$

$3.09 = (3)^y \implies y \approx 1$

To this point, $\mathrm{Rate} = k \cdot [A]^x [B]^1$

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for $\Delta \mathrm{Rate} = \Delta [A]^x$

$4= (2)^x \implies x = 2$

the rate law is

Rate = k·[A]²[B]

6 0

3 years ago

When a substance such as ice melts, its temperature increases. Describe what happens to the arrangement of the water molecules a

gogolik [260]

Answer:

The molecules absorb heat and acquire more kinetic energy.

Explanation:

In a solid, the solids only vibrate about their mean positions but do not translate. When energy is supplied to the molecule in the form of heat, the molecules vibrate faster. Eventually, they acquire sufficient energy to leave their mean positions and translate. Hence the solid crystal collapses.

When ice is heated, water molecules acquire sufficient kinetic energy to translate. The intermolecular bonds are gradually broken in the solid framework as heat is absorbed. The heat required for this is known as the latent heat of fusion.

The temperature remains constant until phase transition is over, then temperature rise resumes.

7 0

4 years ago

Read 2 more answers