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mario62 [17]
3 years ago
5

In the reaction of aluminum hydroxide and caldium nitrate, how many grams of calcium hydroxide will be formed if 12.55 g of alum

inum hydroxide are used with excess calcium nitrate?
Chemistry
1 answer:
Dahasolnce [82]3 years ago
7 0

Answer:

17.76g

Explanation:

We need to write a balanced chemical equation for the reaction:

2Al(OH)3 + 3Ca(NO3)2 ——> 3Ca(OH)2 + 2Al(NO3)3

In the reaction above, it can be seen that 2 moles of aluminum hydroxide yielded 3 moles of calcium hydroxide. This is the theoretical viewpoint.

Now we need to know what actually happened. We need to calculate the actual number of moles of aluminum hydroxide reacted l. We can get this by dividing the mass by the molar mass.

The molar mass of aluminum hydroxide is 27+ 3( 16+1)

= 27 + 51 = 78g/mol

The number of moles is thus: 12.55/78 = 0.16 moles

Now if 2 moles of aluminum hydroxide gave 3 moles of calcium hydroxide, 0.16moles will give : (0.16*3)/2 = 0.24moles

Now we can calculate the mass of calcium hydroxide formed. The mass of calcium hydroxide formed is the number of moles multiplied by the molar mass.

The molar mass of calcium hydroxide is; 40 + 2(17) = 74g/mol

The mass is thus =74 * 0.24 = 17.76g

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Which of the following elements is the least reactive metal? Sodium, Rubidium, Chlorine, and Magnesium.​
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First off chlorine is not a metal so you can ignore that one.

Sodium and Rubidium are in group 1 of the periodic table and Magnesium is in group 2.

Group one metals are more reactive than group two because it is harder for the group two metals to lose their 2 valence (outer most) electrons.

As you go down group 1 there is an increase in the reactivity this is because as you go down there is an increase in the atomic radius which leads to more shielding. This weakens the electrostatic forces of attraction making it easier to lose the outermost electrons, therefore they are more reactive.
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In potassium-argon dating, how does the proportion of remaining potassium in a rock or fossil change over time?
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Consider the reaction. 2 HBr(g) ¡ H2(g) + Br2(g) a. Express the rate of the reaction in terms of the change in concentration of
Studentka2010 [4]

Answer :

(A) The rate expression will be:

Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}

(B) The average rate of the reaction during this time interval is, 0.00176 M/s

(C) The amount of Br₂ (in moles) formed is, 0.0396 mol

Explanation :

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The given rate of reaction is,

2HBr(g)\rightarrow H_2(g)+Br_2(g)

The expression for rate of reaction :

\text{Rate of disappearance of }HBr=-\frac{1}{2}\frac{d[HBr]}{dt}

\text{Rate of disappearance of }H_2=+\frac{d[H_2]}{dt}

\text{Rate of formation of }Br_2=+\frac{d[Br_2]}{dt}

<u>Part A:</u>

The rate expression will be:

Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}

<u>Part B:</u>

\text{Average rate}=-\frac{1}{2}\frac{d[HBr]}{dt}

\text{Average rate}=-\frac{1}{2}\frac{(0.512-0.600)M}{(25.0-0.0)s}

\text{Average rate}=0.00176M/s

The average rate of the reaction during this time interval is, 0.00176 M/s

<u>Part C:</u>

As we are given that the volume of the reaction vessel is 1.50 L.

\frac{d[Br_2]}{dt}=0.00176M/s

\frac{d[Br_2]}{15.0s}=0.00176M/s

[Br_2]=0.00176M/s\times 15.0s

[Br_2]=0.0264M

Now we have to determine the amount of Br₂ (in moles).

\text{Moles of }Br_2=\text{Concentration of }Br_2\times \text{Volume of solution}

\text{Moles of }Br_2=0.0264M\times 1.50L

\text{Moles of }Br_2=0.0396mol

The amount of Br₂ (in moles) formed is, 0.0396 mol

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