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LenKa [72]
3 years ago
6

HELP.....From the information presented in the activity series of metals below (4 metals included), answer the following questio

n.
In a single replacement reaction, would aluminum replace magnesium to form a new compound or would there be no reaction?

------------------------------------

magnesium (most reactive)

aluminum

zinc

iron (least)

a)there would be no reaction

b) aluminum would replace magnesium to form a new compound
Chemistry
1 answer:
san4es73 [151]3 years ago
6 0

Answer:

a)there would be no reaction

Explanation:

The activity series of metals has many functions. The one applicable to this problem is that it can be used to determine whether a reaction will occur or not. Also, based on the positions of metals in the series, we can know how reactive a metal is compared to another.

In a single displacement reaction, a metal replaces another metal based on their position on the activity series. Metals that are higher in the series are generally more reactive than others below them and so will displace them.

 

Would aluminum replace magnesium to form a new compound or would there be no reaction?

Magnesium is higher than aluminum in the activity series. Therefore it is more reactive than aluminum. No reaction will occur.

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Answer:

\large \boxed{\text{2.20 g Pb}}

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

            2PbS   +   3O₂   ⟶  2Pb   +   2SO₃

m/g:      2.54        1.88

2. Calculate the moles of each reactant

\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

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