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3 years ago

How many moles of H2SO4 are needed to completely react with 15.0 mole of Al?

Chemistry

1 answer:

mezya [45]3 years ago

8 0

Answer:

The answer to your question is 22.5 moles of H₂SO₄

Explanation:

Data

moles of H₂SO₄ = ?

moles of Al = 15

Balanced chemical reaction

2Al + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 3H₂

To solve this problem use proportions and cross multiplication. Use the coefficients of the balanced chemical reaction.

2 moles of Al ------------------- 3 moles of H₂SO₄

15 moles of Al ------------------ x

x = (15 x 3) / 2

x = 45/2

x = 22.5 moles of H₂SO₄

-Conclusion

22.5 moles of H₂SO₄ react completely with 15 moles of Al.

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A lake in California contains about 8,500,000,000,000 gallons of water. How can this

aniked [119]

Answer:

8.5 * 10^11

Explanation:

8 0

2 years ago

Gas has a volume of 247.3 ML and is at 100 Celsius and 745 Hg. If the mass of the gas is 0.347 g what is the molar mass of the v

Ahat [919]

Answer:

The molar mass of the vapor is 43.83 g/mol

Explanation:

Given volume of gas = V = 247.3 mL = 0.2473 L

Temperature = T = 100 $^{\circ}C$ = 373 K

Pressure of the gas = P = 745 mmHg (1 atm = 760 mmHg)

$P = \displaystyle \frac{745}{760} \textrm{ atm} = 0.9802 \textrm{ atm}$

Mass of vapor = 0.347 g

Assuming molar mass of gas to be M g/mol

The ideal gas equation is shown below

$\textrm{PV} =\textrm{nRT} \\\textrm{PV} = \displaystyle \frac{m}{M}\textrm{ RT } \\0.98026 \textrm{ atm}\times 0.2473 \textrm{ L} = \displaystyle \frac{3.47 \textrm{ g}}{M}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 373\textrm{K} \\M = 43.834 \textrm{ g/mol}$

The molar mass of the vapor comes out to be 43.834 g/mol

4 0

2 years ago

Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe

baherus [9]

Answer:

The equilibrium constant is $K =0.02867$

The equilibrium pressure for oxygen gas is $P_{O_2} = 0.09367\ atm$

Explanation:

From the question we are told that

The equation of the chemical reaction is

$M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}$

The Gibbs free energy for $M_3 O_4_{(s)}$ is $\Delta G^o_{1} = -8.80 \ kJ/mol$

The Gibbs free energy for $M{(s)}$ is $\Delta G^o_{2} = 0 \ kJ/mol$

The Gibbs free energy for $O_2{(s)}$ is $\Delta G^o_{3} = 0 \ kJ/mol$

The Gibbs free energy of the reaction is mathematically represented as

$\Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r$

$\Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)$

Substituting values

From the balanced equation

$\Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]$

$\Delta G^o_{re} = 8.80 kJ/mol =8800J/mol$

The Gibbs free energy of the reaction can also be represented mathematically as

$\Delta G^o_{re} = -RTln K$

Where R is the gas constant with a value of $R = 8.314 J/mol \cdot K$

T is the temperature with a given value of $T = 298 K$

K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction is mathematically represented as

$K_p =[ P_{O_2}]^{\frac{3}{2} }$

Where $[ P_{O_2}]$ is the equilibrium pressure of oxygen

The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the second equation of the Gibbs free energy of the reaction

$K = e^{- \frac{\Delta G^o_{re}}{RT} }$

Substituting values

$K= e^{\frac{8800}{8.314 * 298} }$

$K =0.02867$

Now substituting this into the equation above to obtain the equilibrium of oxygen

$0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}$

multiplying through by $1 ^{\frac{2}{3} }$

$P_{O_2} = [0.02867]^{\frac{2}{3} }$

$P_{O_2} = 0.09367\ atm$

3 0

3 years ago

Which of the following best summarizes what happens in a displacement reaction?

pav-90 [236]

Answer:

Explanation:

Here's an example of displacement reaction:

2 Al (s) + 3 CuSO4 (aq) -> 3 Cu (s) + Al2(SO4)3 (aq)

note that aluminium and copper are elements,

copper II sulphate and aluminium sulphate are compound

3 0

3 years ago

Select the correct answer.

Sladkaya [172]

Answer:

Option D. There will be a shift toward the reactants.

Explanation:

The reaction is:

$PCl_5(g)+heat\rightleftharpoons PCl_3(g)+Cl_2(g)$

The application of LeChatelier's principle leads to consider the heat as a reactant or a product depending on if it is on the left side or the right side.

In this reaction, the heat is on the left side, thus it must be considered a reactant.

Decreasing the temperature is equivalent to remove or consume heat. Thus, the reaction must shif to the left to compensate that reduction of heat. That is the reverse reaction shall be favored.

In conclusion, there will be a shift toward the reactants.

5 0

3 years ago