Question

A 6.53 g sample of a mixture of magnesium carbonate and calcium carbonate is treated with excess hydrochloric acid. The resulting reaction produces 1.73 L of carbon dioxide gas at 26 ∘C and 745 torr pressure.
a)Calculate the total number of moles of carbon dioxide that forms from these reactions.
Express your answer using three significant figures.
b) Assuming that the reactions are complete, calculate the percentage by mass of magnesium carbonate in the mixture.

Answer 1

Answer:

0.069moles

45.65%

Explanation:

Firstly, we need to calculate the number of moles of CO2 produced. We can use the ideal gas equation for this.

PV = nRT

n = PV/RT

according to the question,

P = 746torr

V = 1.73L

T = 26 = 26 + 273.15 = 299.15K

n = ?

R = 62.364 L. Torr/k.mol

n = (746 * 1.73)/(62.364 * 299.15) = 0.069moles

B. To get this, we can use their molar masses. The molar mass of calcium carbonate is 100g/mol while for magnesium carbonate, molar mass is 84g/mol

The percentage by mass is (84)/(84 + 100) * 6.53g = 2.98g

= 2.98/6.53 * 100 = 45.65%

Answer 2

Given the data from the question,

A. The number of mole of carbon dioxide produced is 0.069 mole

B. The percentage of magnesium carbonate in the mixture is 45.6%

A. How to determine the mole of carbon dioxide

• Volume (V) = 1.73 L
• Pressure (P) = 745 torr
• Temperature (T) = 26 °C = 26 + 273 = 299 K
• Gas constant (R) = 62.364 torr.L/Kmol
• Number of mole (n) =?

Using the ideal gas equation, the mole of carbon dioxide produced can be obtained as follow

PV = nRT

n = PV / RT

n = (745 × 1.73) / ( 62.364 × 299)

n = 0.069 mole

B. How to determine the percentage of MgCO₃

We'll begin by calculating the mass of MgCO₃ in the mixture. This can be obtained as follow:

• Molar mass MgCO₃ = 24 + 12 + (16×3) = 84 g/mol
• Molar mass of CaCO₃ = 40 + 12 + (3×16) = 100 g/mol
• Total = 84 + 100 = 184 g/mol
• Mass of mixture = 6.53 g

• Mass of MgCO₃ =?

Mass of MgCO₃ = (84 / 184) × 6.53

Mass of MgCO₃ = 2.98 g

Finally, we shall determine the percentage of MgCO₃. This can be obtained as follow:

• Mass of mixture = 6.53 g
• Mass of MgCO₃ = 2.98 g

• Percentage of MgCO₃ =?

Percentage of MgCO₃ = (2.98 / 6.53) × 100

Percentage of MgCO₃ = 45.6%

Learn more about ideal gas equation:

https://brainly.com/question/4147359