# Write the balanced equation for the reaction of aluminum with copper(II) sulfate solution. What is the mole ratio of the reactants to one another and what is the mole ratio of Al metal to Cu Metal?

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#### Explanation

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#### Explanation:

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Mr. K Share
May 5, 2017

Look the explanation below :)

#### Explanation:

Aluminium is more electropositive than copper, so it will replace the copper from its salt solution.

So, the ionic formula is of Al is $A {l}^{3 +}$; Cu is $C {u}^{2 +}$ and $S {O}_{4}^{2 -}$

the ratio of oxidation number of aluminium ion to copper is 3:2 so,

1. $2 A l + 3 C u S {O}_{4} \to A {l}_{2} {\left(S {O}_{4}\right)}_{3} + 3 C u$

2. This reaction told us that when 2 moles of $A l$ react with 3 moles of $C u S {O}_{4}$ , you will get 1 mole of aluminium sulfate and 3 moles of copper.
$\therefore$So the ratio of $A l$ metal to $C u$ metal, in terms of moles, is 2:3.

3. From the equation, 2 moles of $A l$ produce 3 moles of $C u$.
In this reaction $x$ moles of $A l$ produce 0.45 moles of $C u$.
$2 : 3 = x : 0.45$
$\frac{2}{3} = \frac{x}{0.45}$
$x = 0.3 m o l$
$\therefore$ You've needed 0.3 moles of $A l$ to produce 0.45 moles of Cu.

4. Molar mass of $C u$ is $63.546 g \cdot m o {l}^{-} 1$. (Normally we use $64 g \cdot m o {l}^{-} 1$ )
Number of moles, $m o l$ = (mass of an element, $g$ ) / (molar mass of that element, $g \cdot m o {l}^{-} 1$)
Number of moles of Cu = $\frac{4.5 g}{64 g \cdot m o {l}^{-} 1}$ = 0.070 moles.
From the equation, 2 moles of $A l$ produce 3 moles of $C u$.
In this reaction $y$ moles of $A l$ produce 0.07 moles of $C u$.
$2 : 3 = y : 0.07$
$\frac{2}{3} = \frac{y}{0.07}$
$y = 0.04667 m o l$
You've needed 0.04667 moles of Aluminium. But what is the mass of 0.04667 moles of $A l$, please continue reading.
mass of an element, $g$ = moles of that element, $m o l$ MULTIPLY its molar mass, $g \cdot m o {l}^{-} 1$
Molar mass of $A l$ = 27 $g \cdot m o {l}^{-} 1$
mass of $A l$ required = $0.04667 m o l \cdot \left(27 g \cdot m o {l}^{-} 1\right)$ = 1.26$g$
$\therefore$ You've required 1.26$g$ of $A l$

5. Copper metal, $C u$ doesn't appear at all before the experiment. The question doesn't make sense.

I hope my explanation is clear for you. :)

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