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2 years ago

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How many moles of \ce{Fe2O3}FeX 2 OX 3 will be produced from 27.0 \text{ g}27.0 g27, point, 0, start text, space, g, end tex

t of \ce{Fe}FeF, e, assuming \ce{O2}OX 2 is available in excess?

1 answer:

ira [324]2 years ago

6 0

Answer : The number of moles of $Fe_2O_3$ produced will be, 0.241 moles.

Solution : Given,

Mass of Fe = 27.0 g

Molar mass of $Fe$ = 56 g/mole

First we have to calculate the moles of $Fe$ .

$\text{ Moles of }Fe=\frac{\text{ Mass of }Fe}{\text{ Molar mass of }Fe}=\frac{27.0g}{56g/mole}=0.482moles$

Now we have to calculate the moles of $Fe_2O_3$

The balanced chemical reaction is,

$4Fe+3O_2\rightarrow 2Fe_2O_3$

From the reaction, we conclude that

As, 4 mole of $Fe$ react to give 2 mole of $Fe_2O_3$

So, 0.482 moles of $Fe$ react to give $\frac{0.482}{4}\times 2=0.241$ moles of $Fe_2O_3$

Thus, the number of moles of $Fe_2O_3$ produced will be, 0.241 moles.

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The concentration of a saturated BaCl2 solution is 1.75 M (mol/liter) and the concentration of a saturated Na2SO4 solution is 2.

Kitty [74]

Answer:

a) The theoretical yield is 408.45g of $BaSO_{4}$

b) Percent yield = $\frac{realyield}{408.45g}*100$

Explanation:

1. First determine the numer of moles of $BaCl_{2}$ and $Na_{2}SO_{4}$ .

Molarity is expressed as:

M= $\frac{molessolute}{Lsolution}$

- For the $BaCl_{2}$

M= $\frac{1.75molesBaCl_{2}}{1Lsolution}$

Therefore there are 1.75 moles of $BaCl_{2}$

- For the $Na_{2}SO_{4}$

M= $\frac{2.0moles[tex]Na_{2}SO_{4}$ }{1Lsolution}[/tex]

Therefore there are 2.0 moles of $Na_{2}SO_{4}$

2. Write the balanced chemical equation for the synthesis of the barium white pigment, $BaSO_{4}$ :

$BaCl_{2}+Na_{2}SO_{4}=BaSO_{4}+2NaCl$

3. Determine the limiting reagent.

To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:

- For the $BaCl_{2}$ :

$\frac{1.75}{1}=1.75$

- For the $Na_{2}SO_{4}$ :

$\frac{2.0}{1}=2.0$

As the $BaCl_{2}$ is the smalles quantity, this is the limiting reagent.

4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.

$1.75molesBaCl_{2}*\frac{1molBaSO_{4}}{1molBaCl_{2}}*\frac{233.4gBaSO_{4}}{1molBaSO_{4}}=408.45gBaSO_{4}$

5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:

Percent yield = $\frac{realyield}{theoreticalyield}*100$

Percent yield = $\frac{realyield}{408.45g}*100$

The real yield is the quantity of barium white pigment you obtained in the laboratory.

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2 years ago

The rate of a certain reaction is given by the following rate law: rate Use this information to answer the question below. At a

AleksAgata [21]

Answer:

Initial rate of the reaction when concentration of hydrogen gas is doubled will be $3.2\times 10^6 M/s$ .

Explanation:

$N_2+3H_2\rightarrow 2NH_3$

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

Initial rate of the reaction = R = $4.0\times 10^5 M/s$

$R = k\times [N_2][H_2]^3$

$4.0\times 10^5 M/s=k\times [N_2][H_2]^3$

The initial rate of the reaction when concentration of hydrogen gas is doubled : R'

$[H_2]'=2[H_2]$

$R'=k\times [N_2][H_2]'^3=k\times [N_2][2H_2]^3$

$R'=8\times k\times [N_2][H_2]^3$

$R'=8\times R=8\times 4\times 10^5 M/s=3.2\times 10^6 M/s$

Initial rate of the reaction when concentration of hydrogen gas is doubled will be $3.2\times 10^6 M/s$ .

7 0

2 years ago

I know the paper is blurry but can anyone explain any of the questions from 76-80?

Whitepunk [10]

Answer:try to make a good pic please I have ideas about the topic

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2 years ago

The equilibrium constant for the gas phase reaction N2 (g) + O2 (g) ⇌ 2NO (g) is Keq = 4.20 ⋅ 10-31 at 30 °C. At equilibrium, __

pychu [463]

Answer:

At equilibrium, reactants predominate.

Explanation:

For every reaction, the equilibrium constant is defined as the ratio between the concentration of products and reactants. Thus, for the reaction N2 (g) + O2 (g) ⇌ 2NO the expression of its equilibrium constant is:

$Keq = \frac{[NO]^{2}}{[O_{2} ][N_{2}]}$

Since the equilibrium constant is Keq = 4.20x10-31 the concentration of reactants O2 and N2 must be much higher than products to obtain such a small number as 4.20x10-31 at the equilibrium. Hence, at equilibrium reactants predominate.

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2 years ago

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slega [8]

Answer:

C.5

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2 years ago