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Elza [17]
4 years ago
14

If the atomic radius of a calcium crystal is 180 pm, what is the length of the edge of the unit cell? (the calcium crystal forms

Chemistry
1 answer:
My name is Ann [436]4 years ago
4 0
<span>Calcium crystal has a face center cubic (FCC) structure. For FCC structure,
relation between edge length (a) and radius (r) is
                                                       a = 2</span>√2 r
<span>Given: r = 180 pm
Therefore, a = 2</span>√2 180
<span>                     = 2 x 1.141 x 180
                     = 410.76 pm

</span><span>The length of the edge of the Ca unit cell is 410.76 pm.</span><span>

 </span>
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Answer:

1. 4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

2. 6 moles of Cl2

Explanation:

1. The balanced equation for the reaction. This is illustrated below:

4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

2. Determination of the number of mole of Cl2 produce when 4 moles of FeCl3 react with 4 moles. To obtain the number of mole of Cl2 produced, we must determine which reactant is the limiting reactant.

This is illustrated below:

From the balanced equation above,

4 moles of FeCl3 reacted with 3 moles of O2.

Since lesser amount of O2 (i.e 3 moles) than what was given (i.e 4 moles) is needed to react completely with 4 moles of FeCl3, therefore FeCl3 is the limiting reactant and O2 is the excess reactant.

Finally, we can obtain the number of mole Cl2 produced from the reaction as follow:

Note: the limiting reactant is used as it will produce the maximum yield of the reaction since all of it is used up in the reaction.

From the balanced equation above,

4 moles of FeCl3 will react to produced 6 moles of Cl2.

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The following reaction shows the products when sulfuric acid and aluminum hydroxide react.
Elden [556K]

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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