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a 3.214 g sample of magnesium reacts with 8.416 g of bromine (br2). the only product is magnesium bromide (mgbr2). how much mgbr2 is formed?

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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