Question

If the original solution contained 0.12 mol hcl, what amount of hcl, in moles, will remain when the reaction is complete? 1.4 g zn reacted.

Answer 1

First, we have to know the reaction between HCl and Zn. This is a single replacement reaction as shown below:

2 HCl + Zn → ZnCl₂ + 2  H⁺

The molar mass for zinc is 65.38 g/mol. Convert mass to moles and compute for moles of HCl needed to completely react.

Mole of Zn = 1.4 g * 1 mol/65.38 g = 0.0214 moles Zn
Moles of HCl reacted =  0.0214 mol Zn * ( 2 mol HCl/1 mol Zn) = 0.0428 mol
Moles HCl remaining = 0.12 - 0.0428 = 0/077 moles HCl

Answer 2

$\boxed{{\text{0}}{\text{.772 mol}}}$ of HCl is remained in the solution when the reaction between Zn and HCl is completed.

Further Explanation:

The five types of chemical reactions are as follows:

1. Combination reactions:

These reactions are also known as a synthesis reaction. These are the reaction in which two or more reactants combine to form a single product. These are generally accompanied by the release of heat so they are exothermic reactions.

Examples of combination reactions are as follows:

(a) ${\text{Ba}}+{{\text{F}}_2}\to{\text{Ba}}{{\text{F}}_2}$

(b) ${\text{CaO}}+{{\text{H}}_2}{\text{O}}\to{\text{Ca}}{\left({{\text{OH}}}\right)_2}$

2. Decomposition reactions:

The opposite of combination reactions is called a decomposition reaction. Here, a single reactant gets broken into two or more products. Such reactions are usually endothermic because energy is required to break the existing bonds between the reactant molecules.

Examples of decomposition reactions are as follows:

(a) $2{{\text{H}}_2}{{\text{O}}_2}\to2{{\text{H}}_2}{\text{O}}+{{\text{O}}_2}$

(b) $2{\text{NaCl}}\to{\text{2Na+C}}{{\text{l}}_2}$

3. Displacement reactions

Also known as replacement or metathesis reactions. Here, one of the reactants gets replaced by the other one. Generally, the more reactive element displaces the less reactive element. Both metals and non-metals can take part in displacement reactions.

Examples of displacement reactions are as follows:

(a) ${\text{Cu}}+{\text{AgN}}{{\text{O}}_3}\to{\text{Ag}}+{\text{Cu}}{\left({{\text{N}}{{\text{O}}_3}}\right)_2}$

(b) ${\text{C}}{{\text{l}}_2}+{\text{KBr}}\to{\text{B}}{{\text{r}}_2}+{\text{KCl}}$

4. Double displacement reactions

These are the reaction in which ions of two compound interchange with each other to form the product. For example, the general double displacement reaction between two compound AX and BY is as follows:

 ${\text{AX}}+{\text{BY}}\to{\text{AY}}+{\text{BX}}$

Examples of double displacement reactions are as follows:

(a) ${\text{N}}{{\text{a}}_2}{\text{S}}+{\text{HCl}}\to{\text{NaCl}}+{{\text{H}}_2}{\text{S}}$

(b) $2{\text{KOH}}+{\text{Cu}}{\left({{\text{N}}{{\text{O}}_{\text{3}}}}\right)_2}\to2{\text{KN}}{{\text{O}}_3}+{\text{Cu}}{\left({{\text{OH}}}\right)_2}$

5. Combustion reactions:

These are the reactions that take place when hydrocarbons are burnt in the presence of oxygen to form carbon dioxide and water. These are also referred to as burning.

Example of combustion reactions are as follows:

(a) ${\text{C}}{{\text{H}}_4}+{{\text{O}}_2}\to{\text{C}}{{\text{O}}_2}+{{\text{H}}_2}{\text{O}}$

(b) ${{\text{C}}_{10}}{{\text{H}}_{14}}+12{{\text{O}}_2}\to10{\text{C}}{{\text{O}}_2}+4{{\text{H}}_2}{\text{O}}$

The given reaction occurs as follows:

$2{\text{HCl}}+{\text{Zn}}\to{\text{ZnC}}{{\text{l}}_2}+2{{\text{H}}^+}$

The given reaction is a displacement reaction.

The formula to calculate the moles of Zn is as follows;

${\text{Moles of Zn}}=\frac{{{\text{Given mass of Zn}}}}{{{\text{Molar mass of Zn}}}}$                         …… (1)

The given mass of Zn is 1.4 g.

The molar mass of Zn is 65.38 g/mol.

Substitute these values in equation (1).

$\begin{gathered}{\text{Moles of Zn}}=\left({{\text{1}}{\text{.4 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{65}}{\text{.38 g}}}}}\right)\\=0.02141\;{\text{mol}}\\\end{gathered}$

In accordance with the given reaction, 1 mole of Zn reacts with 2 moles of HCl.

So the number of moles of HCl that are reacted with 0.0214 moles of Zn are calculated as follows:

$\begin{gathered}{\text{Moles of HCl}}=\left({{\text{0}}{\text{.0214 mol Zn}}}\right)\left({\frac{{{\text{2 mol Zn}}}}{{{\text{1 mol HCl}}}}}\right)\\=0.0428\;{\text{mol}}\\\end{gathered}$

If 1.4 g (0.0214 mol) of Zn is reacted in the reaction, 0.0428 moles of HCl is consumed. But the amount of HCL present in the original solution was 0.12 mol.

The formula to calculate the remaining amount of HCl is as follows:

${\text{Remaining amount of HCl}}=\left({{\text{Initial amount of HCl}}}\right)-\left({{\text{Amount of HCl reacted}}}\right)$       …… (2)

The original amount of HCl in the solution is 0.12 mol.

The amount of HCl reacted in the reaction is 0.0428 mol.

Substitute these values in equation (2).

$\begin{gathered}{\text{Remaining amount of HCl}}={\text{0}}{\text{.12 mol}}-{\text{0}}{\text{.0428 mol}}\\={\text{0}}{\text{.0772 mol}}\\\end{gathered}$

Learn more:

1. Balanced chemical equation: brainly.com/question/1405182

2. Oxidation and reduction reaction: brainly.com/question/2973661

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: Zn, HCl, H+, ZnCl2, double displacement reaction, combination reaction, combustion reaction, displacement reaction, decomposition reaction, amount of HCl, moles, moles of Zn, given mass, molar mass, original solution,