what is the percentage of water in the hydrate (Na2Co3)(10H20)? step by step and solution please? big quiz today!!
step1. find the mol mass of Na2CO3
2) find the mol mass of H2O
percent water= [10*molmassH2O / (molmassNa2CO2 +10*molmassH2O) ] *100
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To find the percentage of water in the hydrate (Na2Co3)(10H20), follow these steps:
Step 1: Find the molar mass of Na2CO3 (sodium carbonate):
- The atomic mass of sodium (Na) is 22.99 g/mol.
- There are two sodium atoms, so multiply the atomic mass by 2: 22.99 g/mol * 2 = 45.98 g/mol.
- The atomic mass of carbon (C) is 12.01 g/mol.
- The atomic mass of oxygen (O) is 16.00 g/mol.
- There are three oxygen atoms, so multiply the atomic mass by 3: 16.00 g/mol * 3 = 48.00 g/mol.
- Add up the atomic masses: 45.98 g/mol + 48.00 g/mol + 16.00 g/mol = 109.98 g/mol.
Step 2: Find the molar mass of H2O (water):
- The atomic mass of hydrogen (H) is 1.01 g/mol.
- There are two hydrogen atoms, so multiply the atomic mass by 2: 1.01 g/mol * 2 = 2.02 g/mol.
- The atomic mass of oxygen (O) is 16.00 g/mol.
- Add up the atomic masses: 2.02 g/mol + 16.00 g/mol = 18.02 g/mol.
Step 3: Calculate the percentage of water:
- Multiply the molar mass of H2O by 10 (since there are 10 H2O molecules): 18.02 g/mol * 10 = 180.20 g/mol.
- Plug the values into the formula: [10 * (molmassH2O) / (molmassNa2CO2 + 10 * molmassH2O)] * 100.
- Replace molmassH2O with 180.20 g/mol and molmassNa2CO2 with 109.98 g/mol: [10 * 180.20 g/mol / (109.98 g/mol + 10 * 180.20 g/mol)] * 100.
- Simplify the equation: [1802 g/mol / (1099.8 g/mol)] * 100.
- Divide the numerator by the denominator: 1802 g/mol / 1099.8 g/mol = 1.6389.
- Multiply by 100 to get the percentage: 1.6389 * 100 = 163.89%.
Therefore, the percentage of water in the hydrate (Na2Co3)(10H20) is approximately 163.89%.