Question

What is the mass of 3 moles of potassium nitrate [KNO3]?
What is the mass of 0.75 moles of Aluminum Oxide [Al2O3]?
What is the mass of 3.5 moles of silver acetate [AgCH3COO]?
What is the mass of 0.25 moles of Calcium Sulfite [CaSO4]

Answer 1

Answer on Question#48456 - Chemistry, Other

Task:

What is the mass of 3 moles of potassium nitrate [KNO3]?

What is the mass of 0.75 moles of Aluminum Oxide [Al2O3]?

What is the mass of 3.5 moles of silver acetate [AgCH3COO]?

What is the mass of 0.25 moles of Calcium Sulfite [CaSO4]?

Solution:

I. What is the mass of 3 moles of potassium nitrate [KNO3]?

1. Find the molecular mass (M) of potassium nitrate [KNO3]. It is the mass of 1 mole of potassium nitrate [KNO3]. Atomic mass $(a_{m})$ of each chemical element we can find at the Periodic table.

$$M[\mathrm{KNO_3}] = a_m[K] + a_m[N] + 3 \times a_m[O] = 39 + 14 + 3 \times 16 = 101$$

2. Find the mass of 3 moles (v – amount of moles):

$$m = M[\mathrm{KNO_3}] \times v = 101 \times 3 = 303\ (\mathrm{g}). \text{ (Answer)}$$

II. What is the mass of 0.75 moles of Aluminum Oxide [Al2O3]?

1. Find the molecular mass (M) of Aluminum Oxide [Al2O3]. It is the mass of 1 mole of Aluminum Oxide [Al2O3]. Atomic mass $(a_m)$ of each chemical element we can find at the Periodic table:

$$M[\mathrm{Al_2O_3}] = 2 \times a_m[Al] + 3 \times a_m[O] = 2 \times 27 + 3 \times 16 = 102$$

2. Find the mass of 0.75 moles (v – amount of moles):

$$m = M[\mathrm{Al_2O_3}] \times v = 102 \times 0,75 = 76,5\ (\mathrm{g}). \text{ (Answer)}$$

III. What is the mass of 3.5 moles of silver acetate [AgCH3COO]?

1. Find the molecular mass (M) of silver acetate [AgCH3COO]. It is the mass of 1 mole of silver acetate [AgCH3COO]. Atomic mass $(a_m)$ of each chemical element we can find at the Periodic table:

$$M[\mathrm{AgCH_3COO}] = a_m[Ag] + 2 \times a_m[C] + 3 \times a_m[H] + 2 \times a_m[O] = 108 + 2 \times 12 + 3 \times 1 + 2 \times 16 = 167$$

2. Find the mass of 3,5 moles (v – amount of moles):

$$m = M[\mathrm{AgCH_3COO}] \times v = 167 \times 3,5 = 584,5\ (\mathrm{g}). \text{ (Answer)}$$

IV. What is the mass of 0.25 moles of Calcium Sulfite [CaSO4]?

1. Find the molecular mass (M) of Calcium Sulfite [CaSO4]. It is the mass of 1 mole of Calcium Sulfite [CaSO4]. Atomic mass $(a_m)$ of each chemical element we can find at the Periodic table:

$$M[\mathrm{CaSO_4}] = a_m[Ca] + a_m[S] + 4 \times a_m[O] = 40 + 32 + 4 \times 16 = 136$$

2. Find the mass of 0.25 mole (v – amount of moles):

$$m = M \left[ \mathrm{CaSO_4} \right] \times v = 136 \times 0,25 = 34 \text{ (g)}. \text{ (Answer)}$$

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