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Answer :
The mass of SrF₂ that can be prepared from the reaction is 8.67 g. To determine the mass of SrF₂ that can be prepared from the reaction of 8.4 g of Sr(OH)₂ with 3.6 g of HF, we need to use stoichiometry.
First, we need to calculate the moles of Sr(OH)₂ and HF using their respective molar masses. The molar mass of Sr(OH)₂ is 121.6 g/mol (1 Sr atom + 2 O atoms + 2 H atoms), and the molar mass of HF is 20.0 g/mol (1 H atom + 1 F atom).
For Sr(OH)₂:
8.4 g / 121.6 g/mol = 0.069 moles
For HF:
3.6 g / 20.0 g/mol = 0.18 moles
According to the balanced equation, the ratio of moles between Sr(OH)₂ and SrF₂ is 1:1. Therefore, the number of moles of SrF₂ formed will also be 0.069 moles.
Now, we can calculate the mass of SrF₂ using its molar mass, which is 125.6 g/mol (1 Sr atom + 2 F atoms):
0.069 moles × 125.6 g/mol = 8.67 g.
Learn more about stoichiometry refer:
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