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A student is working in a lab at school.He combines 40 grams of magnesium oxide (MgO) with 25 grams of lithium fluoride (LiF). MgO+2LiFarrow MgF_(2)+Li_(2)O After the reaction is complete, he still has some unreacted MgO left over. Which one of the following statements is true? LiP was the limiting reactant because it ran out causing the reaction to stop. MgO was the limiting reactant because there are not enough moles to complete the reaction. MgO was the limiting reactant because there is still some unreacted MgO leftover. LIF was the limiting reactant because you started with less of it.

Question

A student is working in a lab at school.He combines 40
grams of magnesium oxide (MgO) with 25 grams of lithium
fluoride (LiF).
MgO+2LiFarrow MgF_(2)+Li_(2)O
After the reaction is complete, he still has some unreacted
MgO left over. Which one of the following statements is
true?
LiP was the limiting reactant because it ran out
causing the reaction to stop.
MgO was the limiting reactant because there are
not enough moles to complete the reaction.
MgO was the limiting reactant because there is still
some unreacted MgO leftover.
LIF was the limiting reactant because you started
with less of it.

A student is working in a lab at school.He combines 40 grams of magnesium oxide (MgO) with 25 grams of lithium fluoride (LiF). MgO+2LiFarrow MgF_(2)+Li_(2)O After the reaction is complete, he still has some unreacted MgO left over. Which one of the following statements is true? LiP was the limiting reactant because it ran out causing the reaction to stop. MgO was the limiting reactant because there are not enough moles to complete the reaction. MgO was the limiting reactant because there is still some unreacted MgO leftover. LIF was the limiting reactant because you started with less of it.

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WesleyMaster · Tutor for 5 years

Answer

### LiF was the limiting reactant because it ran out causing the reaction to stop.

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## Step 1: Calculate moles of MgO<br />MgO has a molar mass of approximately \(40 \, \text{g/mol}\). Calculate the moles of MgO.<br />\[<br />\text{Moles of MgO} = \frac{\text{Mass of MgO}}{\text{Molar Mass of MgO}} = \frac{40 \, \text{g}}{40 \, \text{g/mol}} = 1 \, \text{mol}<br />\]<br /><br />## Step 2: Calculate moles of LiF<br />LiF has a molar mass of approximately \(26 \, \text{g/mol}\). Calculate the moles of LiF.<br />\[<br />\text{Moles of LiF} = \frac{\text{Mass of LiF}}{\text{Molar Mass of LiF}} = \frac{25 \, \text{g}}{26 \, \text{g/mol}} \approx 0.962 \, \text{mol}<br />\]<br /><br />## Step 3: Determine limiting reactant<br />The balanced equation is<br />\[<br />\mathrm{MgO} + 2 \mathrm{LiF} \rightarrow \mathrm{MgF}_{2} + \mathrm{Li}_2\mathrm{O}<br />\]<br />From the equation, 1 mole of MgO requires 2 moles of LiF. Since we have fewer moles of LiF than required (0.962 mol) to react with 1 mole of MgO, LiF is the limiting reactant.<br /><br />## Step 4: Analyze remaining MgO<br />After the reaction, some MgO is left unreacted. This further confirms that LiF was completely consumed.
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