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3 years ago

What is the mass of 3.77mol of K3N?

Chemistry

1 answer:

AleksAgata [21]3 years ago

5 0

<h3>Answer:</h3>

495 g K₃N

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

3.77 mol K₃N

Step 2: Identify Conversions

Molar Mass of K - 39.10 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol

Step 3: Convert

Set up: $\displaystyle 3.77 \ mol \ K_3N(\frac{131.31 \ g \ K_3N}{1 \ mol \ K_3N})$
Multiply/Divide: $\displaystyle 495.039 \ g \ K_3N$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

495.039 g K₃N ≈ 495 g K₃N

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A 200.0mL closed flask contains 2.000mol of carbon monoxide gas and 2.000mol of oxygen gas at the temperature of 300.0K. How man

max2010maxim [7]

Answer:

There will react 0.400 moles of oxygen.

Explanation:

Step 1: Data given

Volume of the closed flask = 200.00 mL = 0.2 L

Number of moles of CO = 2.000 mol

Number of moles of O2 = 2.000 mol

Temperature = 300.0 K

Pressure decreases with 10%

Step 2: The balanced equation

2CO(g)+O2(g)⟶2CO2(g)

Step 3: Calculate the initial pressure of the flask before the reaction

P = nRT/V

⇒ with n = the number of moles (2.000 moles CO + 2.000 moles O2 = 4.000 moles)

⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the temperature = 300.0K

⇒ V = the volume = 200.0 mL = 0.2 L

P = (4 * 0.08206*300)/0.2

P = 492.36 atm

Step 4: When the pressure is 10 % decreased:

The final pressure = 492.36 - 49.236 = 443.124 atm

Step 5: Calculate the number of moles

n = PV/RT

⇒ with n = the number of moles

⇒ with P = the pressure = 443.124 atm

⇒ V = the volume = 200.0 mL = 0.2 L

⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the temperature = 300.0K

n =(443.124*0.2)/(0.08206*300)

n = 3.6 moles = total number of moles

Step 6: Calculate number of moles

For the reaction :2CO(g) + O₂(g) ⟶ 2CO₂(g)

For each mole of O2 we have 2 moles of CO, to produce 2 moles of CO2

Moles CO = (2 -2X) moles

Moles O2 = (2-X) moles

Moles CO2 = 2X

The total number of moles (4 -X)= 3.6 moles

Where X are moles that react

X = 0.400 moles

There will react 0.400 moles of oxygen.

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