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35.0g Mg reacts with 55.0gBr_(2) to form MgBr_(2) Which is the limiting reagent? Mg(s)+Br_(2)(g)arrow MgBr_(2)(s) Br_(2) MgBr_(2) Mg 12 point

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35.0g Mg reacts with 55.0gBr_(2) to form MgBr_(2) Which is the limiting reagent? Mg(s)+Br_(2)(g)arrow MgBr_(2)(s) Br_(2) MgBr_(2) Mg 12 point

35.0g Mg reacts with 55.0gBr_(2) to form MgBr_(2) Which is the limiting reagent? Mg(s)+Br_(2)(g)arrow MgBr_(2)(s) Br_(2) MgBr_(2) Mg 12 point

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CecilElite · Tutor for 8 years

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The limiting reagent is \( \mathrm{Br_{2}} \).

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## Step 1<br />The limiting reagent in a chemical reaction is the component that is completely consumed and therefore determines when the reaction ends. In this case, Magnesium (Mg) and Bromine (Br2) react together to form MgBr2 according to the balanced equation:<br /><br />### \( \mathrm{Mg(s)} + \mathrm{Br_{2}(g)} \rightarrow \mathrm{MgBr_{2}(s)} \)<br /><br />This equation indicates that 1 mole of Mg reacts with 1 mole of Br2 to form 1 mole of MgBr2.<br /><br />## Step 2<br />To identify the limiting reactant, we calculate the moles of both Mg and Br2. The molar mass of Mg is approximately 24g/mol and Br2 is about 160g/mol.<br /><br />## Step 3<br />Calculate the moles of Mg and Br2:<br /><br />Moles of Mg = \( \frac{35g}{24g/mol} = 1.46 \)<br /><br />Moles of Br2 = \( \frac{55g}{160g/mol} = 0.34 \)<br /><br />## Step 4<br />From the calculation, we can see that we have 0.34 mol of Br2 and 1.46 mol of Mg. This implies that in equimolarization, we have excess Mg. Hence, Bromine is the limiting reagent in this reaction.
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