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Len [333]
3 years ago
6

When magnesium burns in air, it combines with oxygen to form magnesium oxide according to the following equation. What mass in g

rams of magnesium oxide is produced from 2.00 mol of magnesium?
2Mg(s) + O2(g) -> 2MgO(s)
What is the given quantity? ______
What is the unknown quantity? ______
Write the two conversion factors needed to solve the problem.
________ _______
Calculate the molar mass of the unknown using the given quantity and the two conversion factors.
______
Chemistry
1 answer:
Kisachek [45]3 years ago
6 0

Answer: The given quantity is 2 moles of magnesium or 48 g of magnesium.

The unknown quantity is mass of MgO in grams.

Molar mass of MgO = 40 g

Explanation: 2Mg+O_2\rightarrow 2MgO

As can be seen from the chemical equation, 2 moles of magnesium react with 1 mole of oxygen gas to give 2 moles of MgO.

According to mole concept, 1 mole of every substance weighs equal to its molar mass.

Thus 2\times 24= 48 g of magnesium produce=2\times 40= 80 g of MgO.

Molar mass is the mass in grams of 1 mole of the substance.

Thus if 2 moles of MgO weigh= 80 g

1 mole of MgO will weigh=\frac{80}{2}\times 1 =40g

Thus molar mass of MgO is 40 g.

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Suppose that in the synthesis of isoamyl acetate by Fisher esterification, a student began with 6.103 grams of acetic acid and 3
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Answer:

The % yield is 90.68 %

Explanation:

<u>Step 1:</u> Data given

Mass of acetic acid = 6.103 grams

Mass of isoamyl alcohol = 3.728 grams

Mass isoamyl acetate obtained = 4.993 grams

Molar mass acetic acid = 60.05 g/mol

Molar mass of isoamyl alcohol = 88.15 g/mol

Molar mass of isoamyl acetate = 130.19 g/mol

<u>Step 2</u>: The balanced equation

C5H12O + CH3COOH → C7H14O2 +H2O                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           Isoamyl alcohol + acetic acid  ---> Isoamyl acetate + water

<u> </u>

<u>Step 3:</u> Calculate moles of acetic acid

Moles acetic acid = Mass acetic acid / Molar mass acetic acid

Moles acetic acid = 6.103 grams / 60.05 g/mol

Moles acetic acid = 0.1016 moles

<u>Step 4</u>: Calculate moles isoamyl alcohol

Moles isoamyl alcohol = 3.728 grams / 88.15 g/mol

Moles isoamy lalcohol = 0.04229 moles

<u>Step 5</u>: Calculate limiting reactant

The mole ratio is 1:1 so isoamyl alcohol has the smallest number of moles. So isoamyl alcohol is the limiting reactant. It will be completely be consumed (0.04229 moles).

Acetic acid is in excess, there will be consumed 0.04229 moles.

There will remain 0.1016 - 0.04229 = 0.05931‬ moles

<u>Step 6: </u>Calculate moles of isoamyl acetate

For 1 mole acetic acid , we need 1 mole of isoamyl alcohol, to produce 1 mole isoamyl acetate and 1 mole of H2O

For 0.04229 moles isoamyl alcohol consumed, there will be produced 0.04229 moles of isoamyl acetate.

<u>Step 7:</u> Calculate mass of isoamyl acetate

Mass isoamyl acetate = moles isoamyl acetate * molar mas isoamyl acetate

Mass isoamyl acetate = 0.04229 moles * 130.19 g/mol

Mass isoamyl acetate = 5.506 grams = theoretical yield

<u>Step 8</u>: Calcuate % yield

% yield = actual yield / theoretical yield

% yield = (4.993 / 5.506)*100%

% yield = 90.68 %

The % yield is 90.68 %

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Ask Your Teacher A 1.04-mole sample of ammonia at 11.0 atm and 25°C in a cylinder fitted with a movable piston expands against a
RUDIKE [14]

Answer:

Explanation:

a) Volume of the gas nRT / P

= 1,04 X 8.3 X 298 / 11 X 10⁵ m³

= 233.85 x 10⁻⁵ m³

= 233 x 10⁻² L

2.33 L

P₁V₁ / T₁ =P₂V₂/T₂

(11 X 2.33) / 298 = (1 X 24.2) / T

T = 281.37 K

= 8.37 degree

b ) w = p x change in volume

= 10⁵ x ( 24.2 - 2.33 ) x 10⁻³ J

= 21.87 X 10² J

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q  = n x Cp x (25 - 8.37 )

= 1.04 x 35.66x 16.63 J

= 616.65 J

ΔU = Q - W

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3 years ago
Write a half-reaction for the oxidation of the manganese in MnCO3(s) to MnO2(s) in neutral groundwater where the carbonate-conta
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Answer:

The half-reaction for the oxidation of the manganese in MnCO_3(s) to MnO_2(s).:

MnCO_3+2H_2O\rightarrow MnO_2(s)+HCO_3^{-}+3H^+

Explanation:

MnCO_3+H_2O\rightarrow MnO_2(s)+HCO_3^{-}

Let us say the medium in which reaction is taking place is an acidic medium. And balancing in an acidic mediums done as;

Step 1: Balance all the atom beside oxygen and hydrogen atom;

MnCO_3+H_2O\rightarrow MnO_2(s)+HCO_3^{-}

Manganese and carbon are balanced.

Step 2: Balance oxygen atom adding water on the required side:

MnCO_3+H_2O+H_2O\rightarrow MnO_2(s)+HCO_3^{-}

MnCO_3+2H_2O\rightarrow MnO_2(s)+HCO_3^{-}

Step 3: Now balance hydrogen atom by adding hydrogen ion on the required side:

MnCO_3+2H_2O\rightarrow MnO_2(s)+HCO_3^{-}+3H^+

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