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Answer :
molar mass MgS2O3 = 136.43 g/mol
1 mole MgS2O3 ------------ 136.43 g
?? moles MgS2O3 --------- 187 g
187 x 1 / 136.43 => 1.3706 moles
1 mole MgS2O3 ------------ 136.43 g
?? moles MgS2O3 --------- 187 g
187 x 1 / 136.43 => 1.3706 moles
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Final answer:
There are 1.0615 moles of MgSO₄ in 187 g of the compound, using the molar mass of 176.124 g/mol for the calculations.
Explanation:
To calculate the number of moles of MgS₂O₃ in 187 g of the compound, we should follow these steps:
- First, determine the molar mass of MgS₂O₃. Unfortunately, MgS₂O₃ seems to be a typo, as it is not a standard compound. However, based on the information provided, let's assume we're dealing with MgSO₄ (magnesium sulfate), a compound that does exist and whose molar mass was seemingly intended to be communicated.
- Use the given molar mass of 176.124 g/mol for the calculations.
- Apply the formula number of moles = mass (g) / molar mass (g/mol). Plugging in the values gives us:
number of moles = 187 g / 176.124 g/mol = 1.0615 moles of MgSO₄.
Therefore, there are 1.0615 moles in 187 g of the compound presumed to be MgSO₄.
