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3 years ago

1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.

Chemistry

1 answer:

andriy [413]3 years ago

5 0

Answer:

A. $m_{NH_3}^{theo} =1.50gNH_3$

B. $Y=82.2\%$

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

$N_2+3H_2\rightarrow 2NH_3$

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

$n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\ n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3$

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

$m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3$

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

$Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%$

Best regards!

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The overall reaction in a commercial heat pack can be represented as How much heat is released when 4.40 moles of iron are react

vodomira [7]

Answer:

Explanation:

The overall equation for this reaction can be represented as: $4 Fe(s) + 3 O_{2(g) } \to 2Fe_2O_{3(s)} \ \ \ \ \Delta H = -1652 \ kJ$

The first question says:

How much heat is released when 4.40 moles of iron is reacted with excess O₂?

Suppose 1652 kJ of heat is being emitted into the surroundings when four(4) moles of Fe reacted with O₂, therefore;

4.40 moles of Fe reacts with:

$=\dfrac{4.40 \ moles \times 1652 \ kJ}{4 \ moles}$

= 1817.2 kJ of heat will be produced.

The second question says:

How much heat is released when 1.00 mole of $Fe_2O_3$ is produced?

Given that 1652 kJ of heat is being emitted into the surroundings when two(2) moles of $Fe_2O_3$ is produced, therefore;

1.00 moles of $Fe_2O_3$ reacts with:

$=\dfrac{1.00 \ moles \times 1652 \ kJ}{2 \ moles}$

= 826 kJ of heat will be produced.

To the third question; we have:

How much heat is released when 1.60 g iron is reacted with excess O₂?

We need to find the number of moles of iron first.

We know that number of moles = mass/molar mass

Thus, the molar mass of iron = 55.8 g/mol

number of moles of iron = (1.60g) / (55.8 g/mol)

number of moles of iron = 0.02867 mol

Thus; $\dfrac{0.02867\ mol \times 1652 \ kJ }{4 \ mol}$

= 11.84 kJ of heat is released.

The last question says:

How much heat is released when 11.8 g Fe and 1.20 g O₂ are reacted?

Again;

the number of moles of Fe = (11.8g) / (55.8 g/mol) = 0.2114 mole of Fe

Thus; $\dfrac{0.2114\ mol \times 1652 \ kJ }{4 \ mol}$

= 87.31 kJ of heat is released.

On the other hand,

the number of moles of O₂ = (1.20g) / (32 g/mol) = 0.0375 mol of O₂

Thus; $\dfrac{0.0375\ mol \times 1652 \ kJ }{3 \ mol}$

= 20.65 kJ of heat is released

Therefore, when these two(2) reactants reacted with each other, it is just the smaller amount of heat that would be released because oxygen tends to be the limiting reactant.

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Given the following values for the change in enthalpy (deltaH) and entropy (deltaS), which of the following processes can occur

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Answer:

Option A and B

Explanation:

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Delta G = Delta H – T * DS

Substituting the given values, we get –

Delta G = -84 -298 *(125/1000) = -121.25 KJ/mol

Delta G is negative hence the process is spontaneous and will not violate the second law of thermodynamics

(b) DeltaH = -84 kj mol-2 (-20 kcal mol-1), DeltaS = -125j mol-2K-1)(-30 cal mol-1 K-1)

Delta G =-84 -298 *(-125/1000) = -46.75 KJ/mol

Delta G is negative hence the process is spontaneous and will not violate the second law of thermodynamics

Delta G = 84 -298 *(125/1000) = +46.75 KJ/mol

Delta G is positive hence the process is non-spontaneous and will violate the second law of thermodynamics

(d) DeltaH = +84 kj mol-2 (+20 kcal mol-1), DeltaS = +125j mol-2K-1)(-30 cal mol-1 K-1)

Delta G = 84 -298 *(-125/1000) = + 121.25 KJ/mol

Delta G is positive hence the process is non-spontaneous and will violate the second law of thermodynamics

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A civil engineering student is considering buying an electric car. However, the conscious student wants to make sure her carbon

svetlana [45]

Answer:

$\text{An gasoline-powered vehicle has }\\\boxed{\text{five times the carbon footprint}}\text{ of an electric vehicle.}$

Explanation:

1. Miles travelled in an average month

$\text{Miles} = \text{1 mo} \times \dfrac{\text{52 wk}}{\text{12 mo}} \times \dfrac{\text{5 da}}{\text{1 wk}} \times \dfrac{\text{16 mi}}{\text{1 da}} = \text{347 mi}$

2. Using a gasoline powered vehicle

(a) Moles of heptane used

$n = \text{347 mi} \times \dfrac{\text{1 gal}}{\text{36.5 mi}}\times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{679.5 g}}{\text{1L}} \times \dfrac{ \text{1 mol}}{\text{100.20 g}}= \text{244 mol}$

(b) Equation for combustion

C₇H₁₆ + O₂ ⟶ 7CO₂ + 8H₂O

(c) Moles of CO₂ formed

$n = \text{244 mol heptane} \times \dfrac{\text{7 mol CO$_{2}$}}{\text{1 mol heptane}} = \text{1710 mol CO$_{2}$}$

(d) Volume of CO₂ formed

At 20 °C and 1 atm, the molar volume of a gas is 24.0 L.

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3. Using an electric vehicle

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$\text{Theor. Energy} = \text{347 mi} \times \dfrac{\text{1 kWh}}{\text{5.2 mi}} = \text{66.7 kWh theor.}$

(b) Actual energy used

The power station is only 85 % efficient.

$\text{Actual energy used} = \text{66.7 kWh theor.} \times \dfrac{\text{100 kWh actual}}{\text{ 85 kWh theor.}}\times \dfrac{\text{3600 kJ}}{\text{1 kWh}}\\\\ = 2.82\times 10^{5} \text{ kJ}\$

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The heat of combustion of methane is -802.3 kJ·mol⁻¹

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4. Comparison

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