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3 years ago

When 16.35 moles of SI reacts with 11.26 moles of N2, how many moles of SI3N4 are formed

Chemistry

1 answer:

Allushta [10]3 years ago

6 0

Answer:

5.450 mol Si₃N₄

Explanation:

Step 1: Write the balanced equation

3 Si + 2 N₂ ⇒ Si₃N₄

Step 2: Establish the theoretical molar ratio between the reactants

The theoretical molar ratio of Si to N₂ is 3:2 = 1.5:1.

Step 3: Establish the experimental molar ratio between the reactants

The experimental molar ratio of Si to N₂ is 16.35:11.26 = 1.45:1. Comparing both molar ratios, we can see that Si is the limiting reactant.

Step 4: Calculate the moles of Si₃N₄ produced from 16.35 moles of Si

The molar ratio of Si to Si₃N₄ is 3:1.

16.35 mol Si × 1 mol Si₃N₄/3 mol Si = 5.450 mol Si₃N₄

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3 years ago

A gas has a pressure of 8.5atm and occupies 24L at 25∘C. What volume (in liters) will the gas occupy if the pressure is increase

Vesna [10]

The volume (in liters) that the gas will occupy if the pressure is increased to 13.5 atm and the temperature is decreased to 15 °C is 15 L

From the question given above, the following data were obtained:

Initial pressure (P₁) = 8.5 atm

Initial volume (V₁) = 24 L

Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K

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Final temperature (T₂) = 15 °C = 15 + 273 = 288 K

<h3>Final volume (V₂) =? </h3>

The final volume of the gas can be obtained by using the combined gas equation as illustrated below:

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\\\frac{8.5 * 24}{298} = \frac{13.5 * V_{2}}{288}\\\\ \frac{204}{298} = \frac{13.5 * V_{2}}{288}\\\\$

Cross multiply

298 × 13.5 × V₂ = 204 × 288

4023 × V₂ = 58752

Divide both side by 4023

$V_{2} = \frac{58752}{4023}\\\\$

Therefore, the final volume of the gas is 15 L

Learn more: brainly.com/question/25547148

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2 years ago

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4 years ago

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Explanation:

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3 years ago

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

bija089 [108]

<u>Answer:</u> The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u> $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

<u>Equation 2:</u> $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$

We are given:

$K_1=K_b$

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Putting values in above equation, we get:

$K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}$

Hence, the value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

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