# What mass of pure aluminum can be dissolved by 13 ml of 1.6 M KOH?

## Question:

What mass of pure aluminum can be dissolved by {eq}\rm 13\ ml {/eq} of {eq}\rm 1.6\ M\ KOH {/eq}?

## Balanced Chemical Equations:

A balanced chemical equation has a useful feature that it conserves the mass of the substances wherein the number of atoms per element on each side of the equation is equal. With this, the coefficients of each substance lead to the expression of the molar ratio between the substances, which allows us to determine the equivalent amount of one substance with respect to another.

The balanced chemical equation for the reaction is expressed as

{eq}\displaystyle 2 Al + 6 KOH \to 2 KAlO_2 + 2 K_2O + 3 H_2 {/eq}

With this equation, we determine that 2 moles of {eq}\displaystyle Al {/eq} corresponds to 6 moles of {eq}\displaystyle KOH {/eq}, such that we express the number of moles of each substance with the relationship, {eq}\displaystyle 6n_{Al} = 2n_{KOH} {/eq} or {eq}\displaystyle 3n_{Al} = n_{KOH} {/eq}. We determine that {eq}\displaystyle n_{KOH} = M\times V {/eq}, where M = 1.6 M is the given molarity and V = 13 mL = 0.013 L is the given volume in liters.

Moreover, we determine that {eq}\displaystyle n_{Al} = \frac{m}{MW} {/eq}, where m is the mass of aluminum and MW = 26.98 g/mol is its molar mass. We use the given equations and the values for the variables. We proceed with the equation to solve for m.

{eq}\begin{align} \displaystyle 3n_{Al} &= n_{KOH}\\ \frac{3m}{MW} &= M\times V\\ m &= \frac{1}{3}M\times V\times MW\\ &= \frac{1}{3}\times 1.6\ M\times 0.013 \ L\times 26.98\ g/mol\\ &\approx\boxed{\rm 0.19\ g\ Al} \end{align} {/eq} Balanced Chemical Equation: Definition & Examples

from

Chapter 10 / Lesson 18
880K

A chemical equation shows the chemical formulas of substances that are reacting and the substances that are produced. The number of atoms of the reactants and products need to be balanced. In this lesson, we will discuss balancing chemical equations.