How many atoms of calcium are consumed if 7.53 grams of calcium nitrate are produced?
To solve this problem, we need to determine how many calcium atoms are consumed when 7.53 grams of calcium nitrate [Ca(NO₃)₂] are produced.
Step-by-Step Calculation:
- Molar mass of calcium nitrate, Ca(NO₃)₂:
- Ca: 40.08 g/mol
- N: 14.01 g/mol × 2 = 28.02 g/mol
- O: 16.00 g/mol × 6 = 96.00 g/mol
- Total: 40.08 + 28.02 + 96.00 = 164.10 g/mol
- Moles of calcium nitrate: 7.53 g164.10 g/mol=0.0459 mol (approximately)\frac{7.53 \text{ g}}{164.10 \text{ g/mol}} = 0.0459 \text{ mol (approximately)}
- Mole ratio:
- 1 mole of Ca(NO₃)₂ contains 1 mole of calcium atoms.
- So, 0.0459 mol Ca(NO₃)₂ → 0.0459 mol Ca atoms
- Number of calcium atoms: 0.0459 mol×6.022×1023 atoms/mol=2.76×1022 atoms0.0459 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} = 2.76 \times 10^{22} \text{ atoms}
✅ Final Answer:
2.76×1022 atoms of calcium\boxed{2.76 \times 10^{22} \text{ atoms of calcium}}
Explanation (300 words):
To determine how many atoms of calcium are consumed, we start by analyzing the compound calcium nitrate, which has the formula Ca(NO₃)₂. This compound is made up of one calcium ion (Ca²⁺) and two nitrate ions (NO₃⁻). Since only one calcium ion is needed per formula unit of calcium nitrate, the number of calcium atoms corresponds directly to the number of calcium nitrate units produced.
Given 7.53 grams of calcium nitrate, we first calculate the number of moles by dividing the mass by the compound’s molar mass. The molar mass of calcium nitrate is calculated by summing the atomic masses of all atoms in the formula: 1 calcium (40.08 g/mol), 2 nitrogen (2 × 14.01 g/mol), and 6 oxygen (6 × 16.00 g/mol), totaling 164.10 g/mol.
Using this molar mass, we convert grams to moles: 7.53164.10≈0.0459 mol\frac{7.53}{164.10} ≈ 0.0459 \text{ mol}
Since there is 1 calcium atom per unit of Ca(NO₃)₂, there are 0.0459 mol of calcium atoms. To find the number of atoms, we multiply the moles by Avogadro’s number (6.022 × 10²³ atoms/mol), giving: 0.0459×6.022×1023≈2.76×1022 calcium atoms0.0459 \times 6.022 \times 10^{23} ≈ 2.76 \times 10^{22} \text{ calcium atoms}
This number represents the amount of calcium atoms that were required to form 7.53 grams of calcium nitrate.
