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kvv77 [185]
3 years ago
10

The following reaction shows the products when sulfuric acid and aluminum hydroxide react.

Chemistry
1 answer:
Elden [556K]3 years ago
6 0

The correct answer is approximately 11.73 grams of sulfuric acid.

The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.

The balanced equation is:

2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O

Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.

The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g

40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.

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Given the thermochemical equations below, What is the standard heat of formation of CuO(s)? 2 Cu2O(s) + O2(g) ---> 4 CuO(s) ∆
OlgaM077 [116]

-130KJ is the standard heat of formation of CuO.

Explanation:

The standard heat of formation or enthalpy change can be calculated by using the formula:

standard heat of formation of reaction = standard enthalpy of formation of product - sum of enthalpy of product formation

Data given:

Cu2O(s) ---> CuO(s) + Cu(s) ∆H° = 11.3 kJ

2 Cu2O(s) + O2(g) ---> 4 CuO(s) ∆H° = -287.9 kJ

CuO + Cu ⇒ Cu2O (-11.3 KJ)      ( Formation of Cu2O)

When 1 mole Cu20 undergoes combustion 1/2 moles of oxygen is consumed.

Cu20 + 1/2 02 ⇒ 2CuO (I/2 of 238.7 KJ) or 119.35 KJ

So standard heat of formation of  formation of Cu0 as:

Cu + 1/2 02 ⇒ CuO

putting the values in the equation

ΔHf = ΔH1 + ΔH2     (ΔH1 + ΔH2  enthalapy of reactants)

heat of formation = -11.3 + (-119.35)

                            = - 130.65kJ

-130.65 KJ is the heat of formation of CuO in the given reaction.

7 0
3 years ago
Read 2 more answers
Simple Chem question - help!
Lunna [17]

d, one atom of oxygen and two atoms of hydrogen

6 0
4 years ago
On the basis of periodic trends, choose the larger atom from each pair (if possible): match the elements in the left column to t
Ne4ueva [31]

The given question is incomplete. The complete question is as follows.

On the basis of periodic trends, choose the larger atom from each pair (if possible): match the elements in the left column to the appropriate blanks in the sentences on the right. make certain each sentence is complete before submitting your answer.

          Sn, F, Ge, not predictable, Cr

Of Ge or Po, the larger atom is .........

Of F or Se, the larger atom is ..........

Of Sn or I, the larger atom is .........

Of Cr or W, the larger atom is ........

Explanation:

When we move down a group then there occurs an increase in atomic size of the atoms due to increase in the number of electrons.

Ge is a group 14 element which lies in period 4 and Po is a group 16 element that lies in period 6. As polonium is larger in size as compared to germanium.

Fluorine is a group 17 element and lies in period 2. Selenium is a group 16 element lies in 4. Therefore, selenium is larger in size as compared to fluorine.

Sn is a group 14 element that lies in period 5 and I is a group 17 element that lies in period 5. Hence, I is a larger atom.

Cr is a d-block element which lies in period 4 and W is also a d-block element which lies in period 6. Hence, W is larger in size than Cr.

Thus, we can conclude that given blanks are matched as follows.

  • Of Ge or Po, the larger atom is Po.
  • Of F or Se, the larger atom is Se.
  • Of Sn or I, the larger atom is I.
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7 0
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Ms. Crawford gave the following instructions to her science class: • Place an empty water bottle on the table and remove the lid
ehidna [41]
A) chemical - gas produced
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Determine the number of equivalents if a 3.89N solution contains 0.76 L of solution
ValentinkaMS [17]

Answer:

2%

Explanation:

oriented C-2, and (3) the minimizing of the number of ... (2) L. A. Mitscher, J. K. Paul, and L. Goldman,Experientia, 19, 195. (1963). ... SOzCeHiBr)3 in 147 ml. of anhydrous methanol containing 0.37 ... bicarbonate and saturated sodium chloride solution, and dried ... determined in 2% chloroform solution; infrared spectra on.

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