Which statement is true?

How many mole are there in carbon dioxide

Feliz [49]

Answer:

1 mole

Explanation:

8 0

4 years ago

Read 2 more answers

A three-step process for producing molten iron metal from Fe2O3 is: 3Fe2O3 + CO → 2Fe3O4 + CO2 Fe3O4 + CO → 3FeO + CO2 FeO + CO

miskamm [114]

Answer:

There is 100.4652 kg of iron produced from 390 kg of Fe2O3

Explanation:

Step 1: Given data

3Fe2O3 + CO → 2Fe3O4 + CO2

Fe3O4 + CO → 3FeO + CO2

FeO + CO → Fe + CO2

Mass of Fe2O3 = 390 kg = 390000 grams

Molar mass of Fe2O3 = 159.69 g/moles

Step 2: Calculate moles of Fe2O3

Moles of Fe2O3 = mass of Fe2O3 / Molar mass of Fe2O3

Moles of Fe2O3 = 390000 grams / 159.69 g/moles = 2442.2 moles

Step 3: Calculate expected moles of Fe3O4

In the first equation, for 3 moles of Fe2O3 consumed ,we get 2 moles of Fe3O4. The mole ratio is 3:2

This means if we consume 2442.2 moles of Fe2O3, there will be produced 2/3 * 2442.2 = 1628.2 moles of Fe3O4

Since the yield for the step is only 84.6 %

This will be 0.846 * 1628.2 = 1377.4 moles of Fe3O4

Step 4: Calculate expected moles of FeO

In the second equation, for 1 mole of Fe3O4 consumed, there is produced 3 moles of FeO

This means for 1377.4 moles of Fe3O4 consumed, there is 3*1377.4 = 4132.2 moles of FeO produced

Since the yield for the step is only 50.8%

This will be 0.508 * 4132.2 = 2099.2 moles of FeO

Step 5: Calculate expected moles of Fe

In the third equation, for 1 mole of FeO consumed, there is produced 1 mole of Fe.

This means for 2099.2 moles of FeO consumed, there is also 2099.2 moles of Fe produced

Since the yield is only 85.7%

This will be 0.857 * 2099.2 = 1799 moles of Fe

Step 6: Calculate mass of Fe

Mass of Fe = moles of Fe * Molar mass of Fe

Mass of Fe = 1799 moles of Fe * 55.845 g/moles = 100465.2 grams = 100.4652 kg of Fe

There is 100.4652 kg of iron produced from 390 kg of Fe2O3

6 0

4 years ago

Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas

hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of $CS_2$ = 1.20 mole

Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

Now put all the given values in this expression, we get :

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

4 0

3 years ago

Trimethylamine, (CH3)3N is a weak base (Kb = 6.3 × 10–5). What volume of this gas, measured at STP, must be dissolved in 2.5 Lof

9966 [12]

Answer:

8.9L is the volume of the gas that must be dissolved.

Explanation:

For a weak base, we can find [(CH₃)₃N] using the equation:

Kb = [OH⁻] [[(CH₃)₃NH⁺] / [(CH₃)₃N]

As [OH⁻] = [[(CH₃)₃NH⁺] and [OH⁻] = 10^-pOH = 3.16x10⁻³M:

6.3x10⁻⁵ = [3.16x10⁻³M][3.16x10⁻³M] / [(CH₃)₃N]

[(CH₃)₃N] = 0.1587M

As the volume is 2.5L, moles are:

2.5L * (0.1587mol / L) = 0.3968moles

Using:

PV = nRT

We can solve for volume of the gas as follows:

P = 1atm at STP; n = 0.3968moles; R = 0.082atmL/molK; T = 273.15K at STP

V = 0.3968mol*0.082atmL/molK*273.15K/1atm

V = 8.9L is the volume of the gas that must be dissolved.

3 0

3 years ago

Please help, its urgent.

san4es73 [151]

Answer:

Mass of excess reactant left = 179.6 g

Limiting reactant = nitrogen

Mass of ammonia formed = 200.6 g

Explanation:

Given data:

Mass of nitrogen = 165.0 g

Mass of hydrogen = 215.0 g

Limiting reactant = ?

Mass of ammonia formed = ?

Mass of excess reactant left = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 165.0 g/ 28 g/mol

Number of moles = 5.9 mol

Number of moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 215.0 g/ 2 g/mol

Number of moles = 107.5 mol

Now we will compare the moles of ammonia with both reactant.

H₂ : NH₃

3 : 2

107.5 : 2/3×107.5 = 71.7 mol

N₂ : NH₃

1 : 2

5.9 : 2/1×5.9 = 11.8 mol

Less number of moles of ammonia are formed by the nitrogen it will act as limiting reactant.

Mass of ammonia formed:

Mass = number of moles × molar mass

Mass = 11.8 mol × 17 g/mol

Mass = 200.6 g

Mass of hydrogen left:

We will compare the moles of hydrogen and nitrogen.

N₂ : H₂

1 : 3

5.9 : 3/1×5.9 = 17.7 mol

Out of 107.5 moles 17.7 moles of hydrogen react with nitrogen.

Number of moles left unreacted = 107.5 - 17.7 mol = 89.8 mol

Mass of hydrogen left:

Mass = number of moles × molar mass

Mass = 89.8 mol × 2 g/mol

Mass = 179.6 g

5 0

3 years ago