Answer to Question #173067 in General Chemistry for Jeremy

Question

Mole Problems

1. Calculate the mass of 2.5 moles of NaCl.
2. Determine the mass in grams of 0.500 moles of Ca(OH)₂.
3. Calculate the mass of 0.010 moles of NH₃.
4. Determine the mass of 3.0 moles of CO₂.
5. Calculate the number of moles in 168.0 g of HgS.
6. Determine the number of moles in 27.0 g of H₂O.
7. Calculate the number of moles in 50.0 g of NH₃.
8. Calculate the mass of 6.022 x 10²³ particles of CaSO₄.
9. Convert 9.500 x 10²⁴ molecules of CH₄ into a mass in grams.
10. Determine the mass of 3.011 x 10²³ atoms of gold.
11. Calculate the number of molecules in 15.0 grams of CaSO₄.
12. Determine the number of molecules in 100.0 g of KNO₃.
13. Which would have more molecules, 2.0 moles of hydrogen gas, H₂, or 2.0 moles of methane gas, CH₄? Explain your answer.
14. Which would have the larger mass, 2.0 moles of hydrogen gas, H₂, or 2.0 moles of methane gas, CH₄? Explain your answer.

Answer 1

1. m(NaCl) = M×n

m(NaCl) = 58.5×2.5 = 146.25 g

2. m((Ca(OH)₂) = 0.500×74 = 37 g

3. m(NH₃) = 0.010×17 = 0.17 g

4. m(CO₂) = 3.0×44 = 132 g

5. n(HgS) = m/M

n(HgS) = 168.0/233 = 0.72 mole

6. n(H₂O) = 27.0/18 = 1.5 mole

7. n(NH₃) = 50.0/17 = 2.94 mole

8. n(CaSO₄) = N/N_A

n(CaSO₄) = 6.022×10²³/6.022×10²³ = 1 mole

m(CaSO₄) = 136 g

9. n(CH₄) = 9.500×10²⁴/6.022×10²³ = 15.77 mole

m(CH₄) = 15.77×18 = 284 g

10. n(Au) = 3.011×10²³/6.022×10²³ = 0.5 mole

m(Au) = 0.5×197 = 98.5 g

11. n(CaSO₄) = m/M

n(CaSO₄) = 15.0/136 = 0.11 mole

N = n×N_A

N = 0.11× 6.022×10²³ = 0.66×10²³ molecules

12. n(KNO₃) = 100/101 = 0.99 mole

N(KNO₃) = 0.99 × 6.022×10²³ = 5.96×10²³ molecules

13. The number of molecules in a mole is constant and is equal to Avogadro's number: 6.022×10²³.

Therefore, 2.0 mole of H₂ and 2.0 mole CH₄ all contain the same number of molecules which is 6.022×10²³ molecules.

14. m(H₂) = 2.0×2 = 4 g

m(CH₄) = 2.0×18 = 36 g

CH₄ have the larger mass.