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3 years ago

use the equation 2C4H10 +13O2-->8CO2+10H2O to find how many moles of oxygen would react with .425 mol C4H10

2 answers:

6 0

2C4H10 + 13O2 → 8CO2 + 10H2O Since the equation is balanced, we can set up a proportion: 13 moles of O2 react with 2 moles of C4H10x moles of O2 react with 0.425 moles of C4H10 13 → 2x → 0.425 x = 13 * 0.425 / 2 = 2.7625 2.7625 moles of O2 react with 0.425 moles of C4H10

boyakko [2]3 years ago

5 0

Answer: 2.76 moles of $O_2$

Explanation:

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.

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

According to stoichiometry :

2 moles of $C_4H_{10}$ combine with 13 moles of $O_2$

0.425 moles of $C_4H_{10}$ will combine with = $\frac{13}{2}\times 0.425=2.76moles$ of $O_2$

Thus 2.76 moles of $O_2$ would react with 0.425 moles of $C_4H_{10}$

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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.

$V = \text{1710 mol } \times \dfrac{\text{24.0 L}}{\text{1 mol}} \times \dfrac{\text{1 gal}}{\text{3.785 L}} = \textbf{10 800 gal}$

3. Using an electric vehicle

(a) Theoretical energy used

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

(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}\$

(c) Combustion of CH₄

CH₄ + 2O₂ ⟶ CO₂ +2 H₂O

(d) Equivalent volume of CO₂

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

$V= 2.82\times 10^{5}\text{ kJ} \times \dfrac{\text{1 mol methane}}{\text{802.3 kJ}} \times \dfrac{\text{1 mol CO$_{2}$} }{\text{1 mol methane}} \times \dfrac{ \text{24.0 L}}{ \text{1 mol CO$_{2}$}}\\\\ \times \dfrac{\text{1 gal}}{\text{3.875 L}} = \textbf{2180 gal}$

4. Comparison

$\dfrac{V_{\text{gasoline}}}{V_{\text{electric}}} = \dfrac{10800}{2180} = 5.0\\\\ \text{An gasoline-powered vehicle has }\\ \boxed{\textbf{ five times the carbon footprint}}\text{ of an electric vehicle.}$

6 0

3 years ago

The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×

Anuta_ua [19.1K]

Answer:

$Kc=~1.49x10^3^4}$

Explanation:

We have the reactions:

A: $N_2_(_g_) + O_2_(_g_) 2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9$

Our target reaction is:

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)$

We have $NO_(_g_)$ as a reactive in the target reaction and $NO_(_g_)$ is present in A reaction but in the products side. So we have to flip reaction A.

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

Then if we add reactions A and B we can obtain the target reaction, so:

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9$

For the final Kc value, we have to keep in mind that when we have to add chemical reactions the total Kc value would be the multiplication of the Kc values in the previous reactions.

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}$