Answer :
Sure, let's calculate the number of grams in 3.4 moles of Ca(NO3)2 step-by-step:
1. Determine the molar masses of the elements involved:
- Calcium (Ca) has an atomic mass of 40.08 g/mol.
- Nitrogen (N) has an atomic mass of 14.01 g/mol.
- Oxygen (O) has an atomic mass of 16.00 g/mol.
2. Calculate the molar mass of the compound Ca(NO3)2:
- The formula Ca(NO3)2 indicates that each molecule contains:
- 1 calcium atom (Ca)
- 2 nitrogen atoms (N)
- 6 oxygen atoms (O) [since there are 2 nitrate ions (NO3) and each nitrate ion has 3 oxygen atoms].
- Sum the molar masses of all atoms in the formula:
[tex]\[ \text{Molar mass of Ca(NO3)2} = 40.08 \, \text{(Ca)} + 2 \times (14.01 \, \text{(N)} + 3 \times 16.00 \, \text{(O)}) \][/tex]
[tex]\[ \text{Molar mass of Ca(NO3)2} = 40.08 + 2 \times (14.01 + 48.00) \][/tex]
[tex]\[ \text{Molar mass of Ca(NO3)2} = 40.08 + 2 \times 62.01 \][/tex]
[tex]\[ \text{Molar mass of Ca(NO3)2} = 40.08 + 124.02 \][/tex]
[tex]\[ \text{Molar mass of Ca(NO3)2} = 164.10 \, \text{g/mol} \][/tex]
3. Calculate the mass of 3.4 moles of Ca(NO3)2:
- Use the formula:
[tex]\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \][/tex]
- Plug in the values:
[tex]\[ \text{Mass} = 3.4 \, \text{moles} \times 164.1 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass} = 557.94 \, \text{grams} \][/tex]
So, the number of grams in 3.4 moles of Ca(NO3)2 is 557.94 grams.
1. Determine the molar masses of the elements involved:
- Calcium (Ca) has an atomic mass of 40.08 g/mol.
- Nitrogen (N) has an atomic mass of 14.01 g/mol.
- Oxygen (O) has an atomic mass of 16.00 g/mol.
2. Calculate the molar mass of the compound Ca(NO3)2:
- The formula Ca(NO3)2 indicates that each molecule contains:
- 1 calcium atom (Ca)
- 2 nitrogen atoms (N)
- 6 oxygen atoms (O) [since there are 2 nitrate ions (NO3) and each nitrate ion has 3 oxygen atoms].
- Sum the molar masses of all atoms in the formula:
[tex]\[ \text{Molar mass of Ca(NO3)2} = 40.08 \, \text{(Ca)} + 2 \times (14.01 \, \text{(N)} + 3 \times 16.00 \, \text{(O)}) \][/tex]
[tex]\[ \text{Molar mass of Ca(NO3)2} = 40.08 + 2 \times (14.01 + 48.00) \][/tex]
[tex]\[ \text{Molar mass of Ca(NO3)2} = 40.08 + 2 \times 62.01 \][/tex]
[tex]\[ \text{Molar mass of Ca(NO3)2} = 40.08 + 124.02 \][/tex]
[tex]\[ \text{Molar mass of Ca(NO3)2} = 164.10 \, \text{g/mol} \][/tex]
3. Calculate the mass of 3.4 moles of Ca(NO3)2:
- Use the formula:
[tex]\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \][/tex]
- Plug in the values:
[tex]\[ \text{Mass} = 3.4 \, \text{moles} \times 164.1 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass} = 557.94 \, \text{grams} \][/tex]
So, the number of grams in 3.4 moles of Ca(NO3)2 is 557.94 grams.
