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Jet001 [13]
3 years ago
6

Why is knowledge of the activity series crucial in choosing which metal to build things with?

Chemistry
1 answer:
crimeas [40]3 years ago
5 0

Answer:

activity series is a list of substances ranked in order of relative reactivity.

For example, magnesium metal can displace hydrogen ions from solution. So magnesium is more reactive than hydrogen:

Mg(s) + 2H⁺(aq) → H₂(g) + Mg²⁺(aq)

Zinc metal can also displace hydrogen ions from solution:

Zn(s) + 2H⁺(aq) → H₂(g) + Zn²⁺(aq)

so zinc is also more active than hydrogen.

But magnesium metal can displace zinc ions from solution:

Mg(s) + Zn²⁺(aq) → Zn(s) + Mg²⁺(aq)

so magnesium is more active than zinc.

The activity series including these elements would be Mg > Zn > H.

Chemists have built up a complete activity series in a similar way.

The most active metals are at the top of the table; the least active are at the bottom.

Any metal that is higher in the series will displace a metal that is below it in a

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. When a newspaper is left in direct sunlight for a few days, the paper begins to turn yellow. The yellow
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Veronica is taking an average of 16 height measurements in a brand-new experiment. Which term is this the best example of? repli
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onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
mestny [16]

Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of CaCN_2 = 105.0 g

Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0
3 years ago
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