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1 year ago

Use the equation: 2Mg + O2→ 2MgO

Chemistry

1 answer:

ludmilkaskok [199]1 year ago

6 0

Answer:

3 moles of O2 needed

Explanation:

2 moles of Mg to one mole O2

so 3 moles of O2 needed

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2 years ago

Zn +

saul85 [17]

Answer:

1. Theoretical yield = 2.03g

2. Actual yield 1.89g

Explanation:

Let us write a balanced equation. This is illustrated below:

Zn + 2HCI —> ZnCl2 + H2

Molar Mass of HCl = 1 +35.5 = 36.5g/mol

Mass of HCl from the balanced equation = 2 x 36.5 = 73g

Molar Mass of H2 = 2x1 = 2g/mol

1. From the equation,

73g of HCl produced 2g of H2.

Therefore, 74g of HCl will produce = (74 x 2)/73 = 2.03g

Therefore, theoretical yield = 2.03g

2. %yield = 93%

Theoretical yield = 2.03g

Actual yield =?

%yield = Actual yield /Theoretical yield x100

Actual yield = %yield x theoretical yield

Actual yield = 93% x 2.03 = (93/100)x2.03 = 1.89g

Actual yield =1.89g

4 0

3 years ago

The Moon shines white light on Earth. Is the Moon a good blackbody

Hitman42 [59]

Answer:

C.) No, because the Moon reflects most of the Sun's light rather than

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Explanation:

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2 years ago

48g of 02 produce how many grams of Al2O3

bixtya [17]

Taking into account the reaction stoichiometry, 102 grams of Al₂O₃ are formed when 48 grams of O₂ react.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

4 Al + 3 O₂ → 2 Al₂O₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Al: 4 moles
O₂: 3 moles
Al₂O₃: 2 moles

The molar mass of the compounds is:

Al: 27 g/mole
O₂: 32 g/mole
Al₂O₃: 102 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Al: 4 moles ×27 g/mole= 108 grams
O₂: 3 moles ×32 g/mole= 96 grams
Al₂O₃: 2 moles ×102 g/mole= 204 grams

<h3>Mass of Al₂O₃ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 96 grams of O₂ form 204 grams of Al₂O₃, 48 grams of O₂ form how much mass of Al₂O₃?

$mass of Al_{2} O_{3} =\frac{48 grams of O_{2} x204 grams of Al_{2} O_{3}}{96 grams of O_{2}}$