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3 years ago

If 0.905 mol Al2O3 is produced in the reaction, what mass of Fe in product's

Chemistry

1 answer:

Marrrta [24]3 years ago

3 0

<u>Answer:</u> Amount of iron produced is 101.36 grams.

<u>Explanation:</u>

For the reaction of aluminium and iron oxide, the equation follows:

$2Al+Fe_2O_3\rightarrow Al_2O_3+2Fe$

By Stoichiometry of the reaction,

When 1 mole of aluminium oxide is produced, then 2 moles of iron is also produced.

So, when 0.905 moles of aluminium oxide is produced, then $\frac{ 2}{1}\times 0.905mol=1.81moles$ of iron is also produced.

Now, to calculate the amount of iron produced, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of iron = 56 g/mol

Moles of iron = 1.81 moles

Putting values in above equation, we get:

$1.81mol=\frac{\text{Mass of iron}}{56g/mol}\\\\\text{Mass of iron produced}=101.36g$

Hence, amount of iron produced is 101.36 grams.

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A 31.5 g wafer of pure gold initially at 69.4 ∘C is submerged into 63.4 g of water at 27.4 ∘C in an insulated container.

liubo4ka [24]

Answer: The final temperature of both substances at thermal equilibrium is 301.0 K

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of gold = 31.5 g

$m_2$ = mass of water = 63.4 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of gold = $69.4^oC=342.4K$

$T_2$ = temperature of water = $27.4^oC=300.4K$

$c_1$ = specific heat of gold = $0.129J/g^0C$

$c_2$ = specific heat of water= $4.184J/g^0C$

Now put all the given values in equation (1), we get

$-31.5\times 0.129\times (T_{final}-342.4)=[63.4\times 4.184\times (T_{final}-300.4)]$

$T_{final}=301.0K$

The final temperature of both substances at thermal equilibrium is 301.0 K

4 0

4 years ago

How many moles of water will be generated during the combustion of 0.38 moles of methyl alcohol (CH3OH)? 2CH3OH + 3O2 2CO2 + 4H2

ELEN [110]

You already got the balanced equation. So the ratio of mole number is the ratio of the coefficient. Then the ratio of methyl alcohol and water is 2:4=1:2. The water generated is 0.38*2=0.76 mol.

6 0

3 years ago

Solid Al(NO3)3 is added to distilled water to produce a solution in which the concentration on nitrate, [NO3^-], is 0.10 M. What

VikaD [51]

Answer:

The concentration of nitrate ion $NO_{3}^{-}$ is 0.033 M.

Explanation:

$Al(NO_{3})_{3} \rightarrow Al^{3+} + NO_{3}^{-}$

1 mole of $[Al^{3+}]$ = 3 mole $NO_{3}^{-}$

Let $[NO_{3}^{-}]$ = 3y

then according to above reaction,

$[Al^{3+}]$ = y

$[NO_{3}^{-}]$ = 0.10 M (given)

$[Al^{3+}]$ = ?

3y = 0.10 M

$y = \frac{0.1}{3}$

y = 0.033 M

So ,concentration of $[Al^{3+}]$ = y = 0.033 M

5 0

3 years ago

Seawater is an example of a _______ solution

My name is Ann [436]

Answer:

C. Solid in liquid

Explanation:

Seawater is an example of a solid in liquid solution.

Sea water is made up of:

water
mineral salts
dissolved gases

A liquid solution is always made up of a solute being dispersed within the solvent medium.

The solvent is the liquid or fluid medium.

Such solutions are homogeneous because the solute particles are distributed evenly or uniformly in the solvent.

The solute is usually present in smaller amount compared to the solvent.

3 0

3 years ago

What mass (in g) of potassium chlorate is required to supply the proper amount of oxygen needed to burn 117.3 g of methane

Brilliant_brown [7]

The mass (in g) of potassium chlorate required to supply the proper amount of oxygen needed to burn 117.3 g of methane is 1196.82 g

<h3>Combustion of methane</h3><h3 />

Methane burns in oxygen to produce carbon (iv) oxide and water according to the equation of the reaction below:

CH₄ + 2O₂ ----> CO₂ + 2H₂O

1 mole of methane requires 2 moles of oxygen for complete combustion

1 mole of methane has a mass of 16 g

moles of methane in 117.3 g = 117.3/16 = 7.33 moles of methane

7.33 moles of methane will require 2 * 7.33 moles of oxygen

7.33 moles of methane will require 14.66 moles of oxygen

<h3>Decomposition of potassium chlorate </h3>

The decomposition of potassium chlorate produces oxygen

The equation of the reaction is given below:

2KClO3 → 2KCl + 3O2.

2 moles of potassium chlorate produces 3 moles of oxygen

14.66 moles of oxygen will be produced by 14.66 * 2/3 moles of potassium chlorate

14.66 moles of oxygen will be produced by 9.77 moles of potassium chlorate

1 mole of potassium chlorate has a mass of 122.5

9.77 moles of potassium chlorate has a mass of 1196.82 g

Therefore, the mass (in g) of potassium chlorate required to supply the proper amount of oxygen needed to burn 117.3 g of methane is 1196.82 g

Learn more about mass and molar mass at: brainly.com/question/15476873

7 0

3 years ago