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3 years ago

Need help I don’t understand this at all need to show my work and strategy it’s stoichiometry gram to gram please help

Chemistry

1 answer:

laila [671]3 years ago

4 0

Answer:

781×10¯² g of MgCl₂.

Explanation:

The balanced equation for the reaction is given below:

Mg + 2HCl —> MgCl₂ + H₂

Next, we shall determine the mass of HCl that reacted and the mass of MgCl₂ produced from the balanced equation. This is illustrated below:

Molar mass of HCl = 1 + 35.5

= 36.5 g/mol

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

Molar mass of MgCl₂ = 24 + (35.5×2)

= 24 + 71

= 95 g/mol

Mass of MgCl₂ from the balanced equation = 1 × 95 = 95 g

SUMMARY:

From the balanced equation above,

73 g of HCl reacted to produce 95 g of MgCl₂.

Finally, we shall determine the mass of MgCl₂ produced by the reaction of 6 g of HCl. This can be obtained as follow:

From the balanced equation above,

73 g of HCl reacted to produce 95 g of MgCl₂.

Therefore, 6 g of HCl will react to produce = (6 × 95)/73 = 781×10¯² g of MgCl₂.

Thus, 781×10¯² g of MgCl₂ were obtained from the reaction.

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stiv31 [10]

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3 years ago

What is wrong with the statement below?

pychu [463]

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3 years ago

During the following chemical reaction, 46.3 grams of C3H6O react with 73.2 grams of O2

ra1l [238]

Answer:

a) O2 is the limiting reactant

b) 75.70 grams CO2 (theoretical yield)

c) There remains 12.81 grams of C3H6O

d) The actual yield CO2 is 34.29 grams

Explanation:

Step 1: Data given

Mass of C3H6O = 46.3 grams

Mass of O2 = 73.2 grams

Molar mass of C3H6O = 58.08 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

C3H6O + 4O2 → 3 CO2 + 3H2O

Step 3: Calculate moles C3H6O

Moles C3H6O = mass C3H6O / molar mass C3H6O

Moles C3H6O = 46.3 grams / 58.08 g/mol

Moles C3H6O = 0.793 moles

Step 4: Calculate moles O2

Moles O2 = 73.2 grams / 32 g/mol

Moles O2 = 2.29 moles

Step 5: Calculate limiting reactant

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed. (2.29 moles).

C3H6O is in excess. There will react 2.29/4 = 0.5725 moles C3H6O

There will remain 0.793 - 0.5725 = 0.2205 moles C3H6O

This is 0.2205 moles * 58.08 g/mol = 12.81 grams

Step 6: Calculate moles of CO2

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

For 2.29 moles O2 we need 3/4 * 2.29 = 1.72 moles CO2

This is 1,72 moles * 44.01 g/mol = 75.70 grams CO2

Step 7: Calculate actual yield

% yield = 45.3 % = 0.453 = (actual yield / theoretical yield)

actual yield = 0.453 * 75.70 = 34.29 grams

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Pachacha [2.7K]

called the Avogadro number

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atoms O/1mol O]

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