What is the vapor pressure of carbon disulfide at its normal boiling point?

Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

4 0

3 years ago

A gas at 1.2 atm has a

Papessa [141]

Answer:

1.26 mole

Explanation:

Using ideal gas law

PV=nRT

R= 0.08206

T=273+87=360

1.2×31=n×0.08206×360

37.2 =n × 29.54

n= 1.26 mole

8 0

3 years ago

How many moles are in 5.03 x 1023 molecules?

AlladinOne [14]

Answer:

Some formulas for calculating mole are

Mole = Mass/ Molar mass

Mole = no of particles / avogadros constant

NB : no of particles can be no of atoms , no of ions , or no of molecules 2. Avogadros number or constant = 6.02 times 10 ^23

so we will be using the second formula

Mole = no of particles / avogadros constant

Mole = 5.03 x 10 ^23/6.02 x10^23

Mole = 8.355x10^45

hope it helps :)

Explanation:

4 0

3 years ago

If gatorade g series, endurance formula, contains 14 grams of carbohydrate per 8 ounces, how many ounces should a basketball pla

Mashcka [7]

Based on recommended amount of carbohydrate, a basketball player should consume about 17 - 34 ounces of gatorade g series during the hour-long game.

<h3>How many ounces of endurance formula gatorade g series, endurance formula should a basketball player consume during an hour-long game if it contains 14 grams of carbohydrate per 8 ounces?</h3>

Carbohydrates are food substances metabolized easily by the body to produce energy.

Given that the recommended amount of carbohydrate to consume to maintain performance is 30–60 g/h.

Also 14 grams of carbohydrate found in 8 ounces of the drink.

30 g of carbohydrate will be present in 30 × 8/14 = 17.1 ounces of gatorade g series

60 g of carbohydrate will be present in 60 × 8/14 =34.3 ounces of gatorade g series.

Therefore, a basketball player should consume about 17 - 34 ounces of gatorade g series during the hour-long game.

Learn more about carbohydrates at: brainly.com/question/797978

8 0

2 years ago

A

MArishka [77]

Answer:

B. bias

Explanation:

7 0

2 years ago

Read 2 more answers