Final answer:
To create 1.00 cup of water from BaCl₂ hydrate would require approximately 1387 grams of the hydrate, calculated using dimensional analysis based on the mole ratio and molar mass of BaCl₂ and H₂O.
Step-by-step explanation:
The question asks how much of BaCl₂ hydrate is needed to produce 1.00 cup of water using dimensional analysis. Given the information that 0.0046 moles of BaCl₂ yield 0.0094 moles of H₂O, and the mole ratio is 2 moles of H₂O per 1 mole of BaCl₂, we can find out the mass of BaCl₂ needed.
To begin, we know that 1 cup of water weighs approximately 240 grams since the density of water is about 1 g/mL and 1 cup is about 240 mL. Using Example 5.3.4, we convert this mass of water to moles, knowing that the molar mass of water is 18.02 g/mol:
- 240 g H₂O × (1 mol H₂O / 18.02 g H₂O) = 13.32 moles H₂O
This is the amount of water we need to create from the hydrate. Now we use the provided mole ratio of H₂O to BaCl₂:
- 13.32 moles H₂O × (1 mole BaCl₂ / 2 moles H₂O) = 6.66 moles BaCl₂
Finally, we use the molar mass of BaCl₂, which is not provided but can be calculated based on the periodic table values (137.33 g/mol for Ba, 35.45 g/mol for each Cl, totaling about 208.23 g/mol for BaCl₂). Therefore:
- 6.66 moles BaCl₂ × (208.23 g BaCl₂ / 1 mole BaCl₂) = 1386.89 g BaCl₂
So, approximately 1387 grams of the BaCl₂ hydrate would be needed to create 1.00 cup of water.