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2 years ago

What mass of sucrose needs to be dissolved in 540 g of water in order to prepare a 15% by mass solution.

Chemistry

1 answer:

Lilit [14]2 years ago

5 0

Answer:

95.3g

Explanation:

Given parameters:

Mass of water = 540g

Percentage by mass of solution = 15%

Unknown:

Mass of sucrose = ?

Solution;

To solve this problem, we must understand that a solution is made up of solute dissolved in a solvent. A solute is the solid dissolved in another. The solvent is the is dissolving medium;

Percentage by mass of solute in solution = $\frac{Mass of solute}{Mass of solution} x 100$

Mass of solution = mass of water + Mass of sucrose = 540 + M

where M is the mass of sucrose;

therefore;

$\frac{M}{540 + M } x 100$ = 15

Solving for M gives 95.3g

Mass of sucrose is 95.3g

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2 years ago

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3 years ago

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

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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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