Question

question 5. calcium (ca), the fifth most - abundant element on earth, is always found combined with other elements because of its high reactivity. how many moles of calcium are in 525 g ca?
question 6. a certain reaction produces 2.81 g of copper(ii) hydroxide, cu(oh)₂. determine the number of moles produced in the reaction.
question 7. ibuprofen, c₁₃h₁₈o₂, is the active ingredient in many nonprescription pain relievers. its molar mass is 206.31 g/mol. if the tablets in a bottle contain a total of 33 g of ibuprofen, how many moles of ibuprofen are in the bottle?
question 8. a sample of iron(iii) chloride has a mass of 26.29 g. how many moles would this be?
question 9. how many moles are contained in 0.43 g al₂o₃?
question 10. a 508 - g sample of sodium bicarbonate (nahco₃) contains how many moles of sodium bicarbonate?

Explanation:

Question 5

Step1: Get Ca molar mass

Molar mass of Ca = $40.08\ \text{g/mol}$

Step2: Calculate moles of Ca

$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{525\ \text{g}}{40.08\ \text{g/mol}}$

Question 6

Step1: Calculate $\text{Cu(OH)}_2$ molar mass

Molar mass = $63.55 + 2\times(16.00+1.01) = 97.57\ \text{g/mol}$

Step2: Calculate moles of $\text{Cu(OH)}_2$

$\text{Moles} = \frac{2.81\ \text{g}}{97.57\ \text{g/mol}}$

Question 7

Step1: Use given ibuprofen molar mass

Molar mass = $206.31\ \text{g/mol}$

Step2: Calculate moles of ibuprofen

$\text{Moles} = \frac{33\ \text{g}}{206.31\ \text{g/mol}}$

Question 8

Step1: Calculate $\text{FeCl}_3$ molar mass

Molar mass = $55.85 + 3\times35.45 = 162.20\ \text{g/mol}$

Step2: Calculate moles of $\text{FeCl}_3$

$\text{Moles} = \frac{26.29\ \text{g}}{162.20\ \text{g/mol}}$

Question 9

Step1: Calculate $\text{Al}_2\text{O}_3$ molar mass

Molar mass = $2\times26.98 + 3\times16.00 = 101.96\ \text{g/mol}$

Step2: Calculate moles of $\text{Al}_2\text{O}_3$

$\text{Moles} = \frac{0.43\ \text{g}}{101.96\ \text{g/mol}}$

Question 10

Step1: Calculate $\text{NaHCO}_3$ molar mass

Molar mass = $22.99 + 1.01 + 12.01 + 3\times16.00 = 84.01\ \text{g/mol}$

Step2: Calculate moles of $\text{NaHCO}_3$

$\text{Moles} = \frac{508\ \text{g}}{84.01\ \text{g/mol}}$

Answer:

1. Moles of Ca: $13.1\ \text{mol}$
2. Moles of $\text{Cu(OH)}_2$: $0.0288\ \text{mol}$
3. Moles of ibuprofen: $0.16\ \text{mol}$
4. Moles of $\text{FeCl}_3$: $0.1621\ \text{mol}$
5. Moles of $\text{Al}_2\text{O}_3$: $0.0042\ \text{mol}$
6. Moles of $\text{NaHCO}_3$: $6.05\ \text{mol}$