# 2.5*g masses each of hydrogen chloride, and calcium carbonate are reacted. Which is the limiting reagent?

Aug 15, 2017

$C {O}_{2} = 1.1 g r a m s$
$C a C {l}_{2} = 2.78 g r a m s$

#### Explanation:

First, you need to write out the balanced chemical reaction equation. Then you convert the given mass to moles to see which compound “limits” the reaction (the one completely used up before the other one is fully reacted). Finally, you again use the balanced equation to convert the actual moles of reagents into the respective moles of product, and then convert those moles back into masses.

CaCO_3 + HCl → CO_2 + CaCl_2 Basic equation, unbalanced in Cl, H and O. This is in part because it is skipping an intermediate step carbonic acid (${H}_{2} C {O}_{3}$) formation that may then decompose into $C {O}_{2}$ and water (${H}_{2} O$).

CaCO_3 + HCl → H_2CO_3 + CaCl_2${H}_{2} O + C {O}_{2} + C a C {l}_{2}$ NOW we can balance it properly:

CaCO_3 + 2HCl → H_2O + CO_2 + CaCl_2 Now it can be seen that it take TWO moles of $H C l$ to react fully with ONE mole of $C a C {O}_{3}$. How many moles do we have?

$\text{2.5g" (CaCO_3)/(100g/(mol)) = 0.025 "mole}$ ; $\text{2.5g"( HCl)/(36.5g/(mol)) = 0.068 "mole}$ Fully reacting 0.025 moles of $C a C {O}_{3}$ requires 0.050 moles HCl. We have 0.068 available, so $C a C {O}_{3}$ is the * LIMITING REAGENT. *

From our equation, see that our 0.025 mole $C a C {O}_{3}$ will produce 0.025 mole each of ${H}_{2} O , C {O}_{2} , C a C {l}_{2}$ (leaving an excess of 0.018 mole of HCl in solution). Converting these into masses we obtain:
$C {O}_{2} = 0.025 m o l \times 44 \frac{g}{m o l} = 1.1 g r a m s$

$C a C {l}_{2} = 0.025 m o l \times 111 \frac{g}{m o l} = 2.775 g r a m s$

A mass balance will confirm the correct calculations. ${H}_{2} O = 0.025 m o l \times 18 \frac{g}{m o l} = 0.45 g r a m s$ ; $H C l = 0.018 m o l \times 36.5 \frac{g}{m o l} = 0.657 g r a m s$

Original Reagent mass: 2.5g + 2.5g = 5.0g
Final Reaction Mass: 1.1g + 2.78g + 0.45g + 0.66g = 5.0g

Aug 15, 2017

$\text{2.5 g CaCO"_3}$ produces ${\text{1.1 g CO}}_{2}$ and ${\text{2.8 g CaCl}}_{2}$.

$\text{2.5 g HCl}$ could produce $\text{1.5 g CO"_2}$ and ${\text{3.8 g CaCl}}_{2}$ only if there were enough $\text{CaCO"_3}$.

$\text{CaCO"_3}$ is the limiting reagent, also called the limiting reactant.

#### Explanation:

Balanced Equation

$\text{CaCO"_3 + "2HCl}$$\rightarrow$$\text{CO"_2 + "CaCl"_2 + "H"_2"O}$

The basic process will be:

$\text{mass reactant}$$\rightarrow$$\text{mol reactant}$$\rightarrow$$\text{mol product}$$\rightarrow$$\text{mass product}$

First convert the masses of calcium carbonate $\left({\text{CaCO}}_{3}\right)$ and hydrochloric acid $\left(\text{HCl}\right)$ to moles by dividing the given mass by the molar mass of each compound $\text{100.086 g/mol}$ for ${\text{CaCO}}_{3}$, and $\text{36.458 g/mol}$ for $\text{HCl}$. Since the molar mass is a fraction, $\text{g"/"mol}$, divide by multiplying the given mass by the inverse of the molar mass.

2.5color(red)cancel(color(black)("g CaCO"_3))xx(1"mol CaCO"_3)/(100.086color(red)cancel(color(black)("g CaCO"_3)))="0.0250 mol CaCO"_3"

2.5color(red)cancel(color(black)("g HCl"))xx(1"mol HCl")/(36.458color(red)cancel(color(black)("g HCl")))="0.0686 mol HCl"

color(red)("Calcium Carbonate: CaCO"_3"

Theoretical Mass of $\text{CO"_2}$

Multiply mol $\text{CaCO"_3}$ by the mol ratio between $\text{CaCO"_3}$ and $\text{CO"_2}$ from the blanced equation.

0.0250color(red)cancel(color(black)("mol CaCO"_3))xx(1"mol CO"_2)/(1color(red)cancel(color(black)("mol CaCO"_3)))="0.0250 mol CO"_2"

Determine the mass in grams of $\text{CO"_2}$ produced by multiplying the mol ${\text{CO}}_{2}$ by its molar mass $\left(\text{44.009 g/mol}\right)$.

0.0250color(red)cancel(color(black)("mol CO"_2))xx(44.009"g CO"_2)/(1color(red)cancel(color(black)("mol CO"_2)))="1.1 g CO"_2" (rounded to two sig figs due to $\text{2.5 g}$)

Theoretical Mass of $\text{CaCl"_2}$

Multiply the mol $\text{CaCO"_3}$ by the mol ratio between $\text{CaCO"_3}$ and $\text{CaCl"_2}$ from the balanced equation.

0.0250color(red)cancel(color(black)("mol CaCO"_3))xx(1"mol CaCl"_2)/(1color(red)cancel(color(black)("mol CaCO"_3)))="0.0250 mol CaCl"_2"

Determine the mass in grams of $\text{CaCl"_2}$ by multiplying the mol ${\text{CaCl}}_{2}$ by its molar mass $\left(\text{110.978 g/mol}\right)$.

0.0250color(red)cancel(color(black)("mol CaCl"_2))xx(110.978"g CaCl"_2)/(1color(red)cancel(color(black)("mol CaCl"_2)))="2.8 g CaCl"_2 (rounded to two sig figs due to $\text{2.5 g}$)

color(blue)("Hydrochloric Acid: HCl"

Theoretical Mass of $\text{CO"_2}$

Multiply the mol $\text{HCl}$ by the mol ratio between $\text{HCl}$ and $\text{CO"_2}$ from the balanced equation.

0.0686color(red)cancel(color(black)("mol HCl"))xx(1"mol CO"_2)/(2color(red)cancel(color(black)("mol HCl")))="0.0343 mol CO"_2"

Determine the mass in grams of $\text{CO"_2}$ produced by multiplying the mol ${\text{CO}}_{2}$ by its molar mass $\left(\text{44.009 g/mol}\right)$.

0.0343color(red)cancel(color(black)("mol CO"_2))xx(44.009"g CO"_2)/(1color(red)cancel(color(black)("mol CO"_2)))="1.5 g CO"_2" (rounded to two sig figs due to $\text{2.5 g}$)

Theoretical Mass of ${\text{CaCl}}_{2}$

Determine the mol ${\text{CaCl}}_{2}$.

Multiply the mol $\text{HCl}$ by the mol ratio between $\text{HCl}$ and $\text{CaCl"_2}$ from the balanced equation.

0.0686color(red)cancel(color(black)("mol HCl"))xx(1"mol CaCl"_2)/(2color(red)cancel(color(black)("mol HCl")))="0.0343 mol CaCl"_2"

Determine the mass in grams of $\text{CaCl"_2}$ by multiplying the mol ${\text{CaCl}}_{2}$ by its molar mass $\left(\text{110.978 g/mol}\right)$.

0.0343color(red)cancel(color(black)("mol CaCl"_2))xx(110.978"g CaCl"_2)/(1color(red)cancel(color(black)("mol CaCl"_2)))="3.8 g CaCl"_2 (rounded to two sig figs due to "2.5 g")

Summary

$\text{2.5 g CaCO"_3}$ produces ${\text{1.1 g CO}}_{2}$ and ${\text{2.8 g CaCl}}_{2}$.

$\text{2.5 g HCl}$ could produce $\text{1.5 g CO"_2}$ and ${\text{3.8 g CaCl}}_{2}$ only if there were more $\text{CaCO"_3}$. $\text{HCl}$ is present in excess.

$\text{CaCO"_3}$ is the limiting reagent.

Aug 15, 2017

We need (i) a stoichiometric equation.......

#### Explanation:

$C a C {O}_{3} \left(s\right) + 2 H C l \left(a q\right) \rightarrow C a C {l}_{2} \left(a q\right) + 2 {H}_{2} O \left(l\right) + C {O}_{2} \left(g\right) \uparrow$

And (ii) we need equivalent quantities of metal salt, and hydrogen chloride.

$\text{Moles of calcium carbonate} = \frac{2.5 \cdot g}{100.09 \cdot g \cdot m o {l}^{-} 1} = 2.49 \times {10}^{-} 3 \cdot m o l .$

$\text{Moles of hydrogen chloride} = \frac{2.5 \cdot g}{36.46 \cdot g \cdot m o {l}^{-} 1} = 0.0686 \cdot m o l .$

And thus the acid is in VAST stoichiometric excess, and thus calcium carbonate is the LIMITING reagent.

And so we gets.................

$2.49 \times {10}^{-} 3 \cdot m o l \times 110.98 \cdot g \cdot m o {l}^{-} 1 = 0.276 \cdot g$ with respect to $\text{calcium chloride}$.

And likewise we gets...........
$2.49 \times {10}^{-} 3 \cdot m o l \times 44.01 \cdot g \cdot m o {l}^{-} 1 = 0.110 \cdot g$ with respect to $\text{carbon dioxide}$.

An extension of this question would be to ask the VOLUME gas produced under standard conditions of $1 \cdot a t m$ (or whatever), and $298 \cdot K$.