The formula for nickel(II) bromide is NiBr2. 1. How many GRAMS of Ni2+ are present in 2.06 moles of nickel(II) bromide? 2. How many MOLES of Br- ion are present in 2.22 grams of nickel(II) bromide?
The Correct Answer and Explanation is:
- How many grams of Ni²⁺ are present in 2.06 moles of nickel(II) bromide?
To find the number of grams of Ni²⁺ in 2.06 moles of NiBr₂, we need to follow these steps:
- First, recognize that in nickel(II) bromide, NiBr₂, each formula unit contains one Ni²⁺ ion.
- Therefore, the number of moles of Ni²⁺ will be the same as the number of moles of NiBr₂.
- We know that the molar mass of Ni (Nickel) is approximately 58.69 g/mol.
So, to calculate the grams of Ni²⁺: Grams of Ni²⁺=moles of Ni²⁺×molar mass of Ni\text{Grams of Ni²⁺} = \text{moles of Ni²⁺} \times \text{molar mass of Ni}Grams of Ni²⁺=moles of Ni²⁺×molar mass of Ni Grams of Ni²⁺=2.06 moles×58.69 g/mol\text{Grams of Ni²⁺} = 2.06 \, \text{moles} \times 58.69 \, \text{g/mol}Grams of Ni²⁺=2.06moles×58.69g/mol Grams of Ni²⁺=120.92 g\text{Grams of Ni²⁺} = 120.92 \, \text{g}Grams of Ni²⁺=120.92g
Thus, 2.06 moles of nickel(II) bromide contain 120.92 grams of Ni²⁺.
- How many moles of Br⁻ ion are present in 2.22 grams of nickel(II) bromide?
To calculate the number of moles of Br⁻ ions in 2.22 grams of NiBr₂, we need to:
- Find the molar mass of NiBr₂. The molar mass of Ni is 58.69 g/mol, and the molar mass of Br (Bromine) is approximately 79.90 g/mol.
The molar mass of NiBr₂ is: Molar mass of NiBr₂=58.69 g/mol+(2×79.90 g/mol)\text{Molar mass of NiBr₂} = 58.69 \, \text{g/mol} + (2 \times 79.90 \, \text{g/mol})Molar mass of NiBr₂=58.69g/mol+(2×79.90g/mol) Molar mass of NiBr₂=58.69 g/mol+159.80 g/mol=218.49 g/mol\text{Molar mass of NiBr₂} = 58.69 \, \text{g/mol} + 159.80 \, \text{g/mol} = 218.49 \, \text{g/mol}Molar mass of NiBr₂=58.69g/mol+159.80g/mol=218.49g/mol
- Now, calculate the moles of NiBr₂ in 2.22 grams:
Moles of NiBr₂=mass of NiBr₂molar mass of NiBr₂\text{Moles of NiBr₂} = \frac{\text{mass of NiBr₂}}{\text{molar mass of NiBr₂}}Moles of NiBr₂=molar mass of NiBr₂mass of NiBr₂ Moles of NiBr₂=2.22 g218.49 g/mol=0.01016 mol\text{Moles of NiBr₂} = \frac{2.22 \, \text{g}}{218.49 \, \text{g/mol}} = 0.01016 \, \text{mol}Moles of NiBr₂=218.49g/mol2.22g=0.01016mol
- Since there are 2 Br⁻ ions per formula unit of NiBr₂, the number of moles of Br⁻ ions will be twice the number of moles of NiBr₂:
Moles of Br⁻=2×moles of NiBr₂=2×0.01016 mol=0.02032 mol\text{Moles of Br⁻} = 2 \times \text{moles of NiBr₂} = 2 \times 0.01016 \, \text{mol} = 0.02032 \, \text{mol}Moles of Br⁻=2×moles of NiBr₂=2×0.01016mol=0.02032mol
Thus, 2.22 grams of nickel(II) bromide contain 0.02032 moles of Br⁻.
In summary:
- There are 120.92 grams of Ni²⁺ in 2.06 moles of NiBr₂.
- There are 0.02032 moles of Br⁻ in 2.22 grams of NiBr₂.
