## How to Find the Limiting Reagent

Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.

Step Two: Find the GFW of each compound (do not combine them).

Step Three: Find the number of moles for each compound. You can do this by dividing the number of grams for one of the compounds by its GFW (do this for both compounds, separately).

Step Four: Take the ratio and rewrite it so that it’s a one to something ratio (1:?). This is done by dividing both numbers in the ratio by the smaller number in the ratio.

Step Five: Whichever compound is the “one part” of the ratio, multiply its number of moles by the other number in the ratio. If the product is greater than the number of moles in the second compound, the second compound is the limiting reagent. If the product is less than the number of moles in the second compound, the first compound is the limiting reagent.

Step Two: Find the GFW of each compound (do not combine them).

Step Three: Find the number of moles for each compound. You can do this by dividing the number of grams for one of the compounds by its GFW (do this for both compounds, separately).

Step Four: Take the ratio and rewrite it so that it’s a one to something ratio (1:?). This is done by dividing both numbers in the ratio by the smaller number in the ratio.

Step Five: Whichever compound is the “one part” of the ratio, multiply its number of moles by the other number in the ratio. If the product is greater than the number of moles in the second compound, the second compound is the limiting reagent. If the product is less than the number of moles in the second compound, the first compound is the limiting reagent.

## Three Worked Examples of Finding the Limiting Reagent

1)

Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.

Step Two: Find the GFW of each compound (do not combine them).

Step Three: Find the number of moles for each compound. You can do this by dividing the number of grams for one of the compounds by its GFW (do this for both compounds, separately).

Step Four: Take the ratio and rewrite it so that it’s a one to something ratio (1:?). This is done by dividing both numbers in the ratio by the smaller number in the ratio.

Step Five: Whichever compound is the “one part” of the ratio, multiply its number of moles by the other number in the ratio. If the product is greater than the number of moles in the second compound, the second compound is the limiting reagent. If the product is less than the number of moles in the second compound, the first compound is the limiting reagent.

===================================================

2)

Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.

Step Two: Find the GFW of each compound (do not combine them).

Step Three: Step Three: Find the number of moles for each compound. You can do this by dividing the number of grams for one of the compounds by its GFW (do this for both compounds, separately).

Step Four: Take the ratio and rewrite it so that it’s a one to something ratio (1:?). This is done by dividing both numbers in the ratio by the smaller number in the ratio.

Step Five: Whichever compound is the “one part” of the ratio, multiply its number of moles by the other number in the ratio. If the product is greater than the number of moles in the second compound, the second compound is the limiting reagent. If the product is less than the number of moles in the second compound, the first compound is the limiting reagent.

====================================================

3)

Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.

Step Two: Find the GFW of each compound (do not combine them).

Step Three: Find the number of moles for each compound. You can do this by dividing the number of grams for one of the compounds by its GFW (do this for both compounds, separately).

Step Four: Take the ratio and rewrite it so that it’s a one to something ratio (1:?). This is done by dividing both numbers in the ratio by the smaller number in the ratio.

Step Five: Whichever compound is the “one part” of the ratio, multiply its number of moles by the other number in the ratio. If the product is greater than the number of moles in the second compound, the second compound is the limiting reagent. If the product is less than the number of moles in the second compound, the first compound is the limiting reagent.

*For the balanced equation below, what is the limiting reagent of 48g of Mg when it is reacted with 110g of HCl: Mg + 2HCl --> MgCl2 + H2*Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.

**The ratio for this equation is going to be 1:2 [magnesium (Mg): hydrochloric acid (HCl)]**

Step Two: Find the GFW of each compound (do not combine them).

**Mg: HCl:**

Mg - 1x24 = 24 g/mol H – 1x1 = 1 g/mol

Cl- 1x35.5 = 35.5 g/mol

GFW of Mg = 24 g/mol GFW of HCl = 36.5 g/mol

Mg - 1x24 = 24 g/mol H – 1x1 = 1 g/mol

Cl- 1x35.5 = 35.5 g/mol

GFW of Mg = 24 g/mol GFW of HCl = 36.5 g/mol

Step Three: Find the number of moles for each compound. You can do this by dividing the number of grams for one of the compounds by its GFW (do this for both compounds, separately).

**Mg: HCl:**

moles = 48g/24GWF moles = 110g/36.5GWF

= 2 moles =3.01 molesmoles = 48g/24GWF moles = 110g/36.5GWF

= 2 moles =3.01 moles

Step Four: Take the ratio and rewrite it so that it’s a one to something ratio (1:?). This is done by dividing both numbers in the ratio by the smaller number in the ratio.

**For this equation you do not have to rewrite the ratio because its already a**__1__:2 ratioStep Five: Whichever compound is the “one part” of the ratio, multiply its number of moles by the other number in the ratio. If the product is greater than the number of moles in the second compound, the second compound is the limiting reagent. If the product is less than the number of moles in the second compound, the first compound is the limiting reagent.

**For every 1 mole of Mg, you need two moles of HCl. Since the “one part” is Mg which is 2 moles, multiply this by 2 (the second part of the ratio). When you multiply them (2mol x 2) you get that the second compound should have 4 moles for there to be no limiting reagent. Since the second compound only has 3.01 moles it must be the limiting reagent because 3.01 is less than 4. Therefore, HCl is the limiting reagent.**===================================================

2)

*For the balanced equation below, what is the limiting reagent of 36g of Mg when it is reacted with 146g of HCl: Mg + 2HCl --> MgCl2 + H2*Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.

**The ratio for this equation is going to be 1:2 [magnesium (Mg): hydrochloric acid (HCl)]**Step Two: Find the GFW of each compound (do not combine them).

**Mg: HCl:**

Mg - 1x24 = 24 g/mol H – 1x1 = 1 g/mol

Cl- 1x35.5 = 35.5 g/mol

GFW of Mg = 24 g/mol GFW of HCl = 36.5 g/mol

Mg - 1x24 = 24 g/mol H – 1x1 = 1 g/mol

Cl- 1x35.5 = 35.5 g/mol

GFW of Mg = 24 g/mol GFW of HCl = 36.5 g/mol

Step Three: Step Three: Find the number of moles for each compound. You can do this by dividing the number of grams for one of the compounds by its GFW (do this for both compounds, separately).

**Mg: HCl:**

moles = 36g/24GWF moles = 146g/36.5GWF

= 1.5 moles =4 molesmoles = 36g/24GWF moles = 146g/36.5GWF

= 1.5 moles =4 moles

Step Four: Take the ratio and rewrite it so that it’s a one to something ratio (1:?). This is done by dividing both numbers in the ratio by the smaller number in the ratio.

**For this equation you do not have to rewrite the ratio because its already a**__1__:2 ratioStep Five: Whichever compound is the “one part” of the ratio, multiply its number of moles by the other number in the ratio. If the product is greater than the number of moles in the second compound, the second compound is the limiting reagent. If the product is less than the number of moles in the second compound, the first compound is the limiting reagent.

**For every 1 mole of Mg, you need two moles of HCl. Since the “one part” is Mg which is 1.5 moles, multiply this by 2 (the second part of the ratio). When you multiply them (1.5mol x 2) you get that the second compound should have 3 moles for there to be no limiting reagent. Since the second compound has 4 moles it must be an excess because 4 is greater than 3. Therefore, Mg is the limiting reagent because HCl is the excess.**====================================================

3)

*For the balanced equation below, what is the limiting reagent of 96g of Mg when it is reacted with 292g of HCl: Mg + 2HCl --> MgCl2 + H2*Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.

**The ratio for this equation is going to be 1:2 [magnesium (Mg): hydrochloric acid (HCl)]**Step Two: Find the GFW of each compound (do not combine them).

**Mg: HCl:**

Mg - 1x24 = 24 g/mol H – 1x1 = 1 g/mol

Cl- 1x35.5 = 35.5 g/mol

GFW of Mg = 24 g/mol GFW of HCl = 36.5 g/molMg - 1x24 = 24 g/mol H – 1x1 = 1 g/mol

Cl- 1x35.5 = 35.5 g/mol

GFW of Mg = 24 g/mol GFW of HCl = 36.5 g/mol

Step Three: Find the number of moles for each compound. You can do this by dividing the number of grams for one of the compounds by its GFW (do this for both compounds, separately).

**Mg: HCl:**

moles = 96g/24GWF moles = 292g/36.5GWF

= 4 moles =8 molesmoles = 96g/24GWF moles = 292g/36.5GWF

= 4 moles =8 moles

Step Four: Take the ratio and rewrite it so that it’s a one to something ratio (1:?). This is done by dividing both numbers in the ratio by the smaller number in the ratio.

**For this equation you do not have to rewrite the ratio because its already a**__1__:2 ratioStep Five: Whichever compound is the “one part” of the ratio, multiply its number of moles by the other number in the ratio. If the product is greater than the number of moles in the second compound, the second compound is the limiting reagent. If the product is less than the number of moles in the second compound, the first compound is the limiting reagent.

**For every 1 mole of Mg, you need two moles of HCl. Since the “one part” is Mg which is 4 moles, multiply this by 2 (the second part of the ratio). When you multiply them (4mol x 2) you get that the second compound should have 8 moles for there to be no limiting reagent. Since the second compound has 8 moles, both compounds have the number of moles that they should have for there to be no limiting reagent. Therefore, this is a perfect ratio and there is no limiting reagent.**