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Sagot :
The reaction between 3.214 g magnesium and 8.416 g bromine produces 9.7 g of MgBr2.
The balanced chemical reaction of magnesium and bromine is shown below.
Mg + Br2 → MgBr2
Step 1: Convert the given mass of reactants to moles of reactants
moles Mg = mass Mg / molar mass Mg
moles Mg = 3.214 g / 24.31 g/mol
moles Mg = 0.1322 mol
moles Br2 = mass Br2 / molar mass Br2
moles Br2 = 8.416 g / 159.808 g/mol
moles Br2 = 0.0527 mol
Step 2: Bromine (Br2) is determined to be the limiting reactant. Since the stoichiometric ratio between the reactants is equal to 1, the reactant with a lower number of moles would be the limiting reactant.
Step 3: Convert moles of Br2 to the mass of MgBr2. Their stoichiometric ratio is also 1, which makes the solution easier.
mol Br2 = mol MgBr2
mass MgBr2 = mol MgBr2 * molar mass MgBr2
mass MgBr2 = 0.0527 mol * 184.118 g /mol
mass MgBr2 = 9.70 g
To learn more about stoichiometric ratios, please refer to https://brainly.com/question/9413308.
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