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3 years ago

3Al + 3 NH4ClO4 ---> Al2O3 + AlCl3 + 3 NO + 6H20

Chemistry

1 answer:

Tresset [83]3 years ago

6 0

Answer:

9.63 L of NO

Explanation:

We'll begin by calculating the number of mole in 50.0 g of NH₄ClO₄. This can be obtained as follow:

Mass of NH₄ClO₄ = 50 g

Molar mass of NH₄ClO₄ = 14 + (4×1) + 35.5 + (16×4)

= 14 + 4 + 35.5 + 64

= 117.5 g/mol

Mole of NH₄ClO₄ =?

Mole = mass /molar mass

Mole of NH₄ClO₄ = 50/117.5

Mole of NH₄ClO₄ = 0.43 mole

Next, we shall determine the number of mole of NO produced by the reaction of 50 g (i.e 0.43 mole) of NH₄ClO₄. This can be obtained as follow:

3Al + 3NH₄ClO₄ –> Al₂O₃ + AlCl₃ + 3NO + 6H₂O

From the balanced equation above,

3 moles of NH₄ClO₄ reacted to produce 3 moles of NO.

Therefore, 0.43 mole of NH₄ClO₄ will also react to produce 0.43 mole of NO.

Finally, we shall determine the volume occupied by 0.43 mole of NO. This can be obtained as follow:

1 mole of NO = 22.4 L

Therefore,

0.43 mole of NO = 0.43 × 22.4

0.43 mole of NO = 9.63 L

Thus, 9.63 L of NO were obtained from the reaction.

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When one atom of carbon reacts with two atoms of oxygen, they combine to form one molecule of carbon dioxide. How many atoms of

tankabanditka [31]

The equation is: C+O2=>CO2

Since we got 10 molecules of CO2 new balanced equation would be 10C+10O2=>10CO2

from this equation we can see that we have 10 molecules of oxygen, however ,we need to find atoms. There are 2 atoms in the oxygen molecule so we need to multiply 10 by 2 which gives us 20 atoms.

The answer: there are 20 atoms of oxygen

4 0

3 years ago

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2

Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

$Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}$

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

$volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL$

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

7 0

3 years ago

If 5.100 g of c6h6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °c, what is the final temp

emmasim [6.3K]

C6H12 = 6x12 + 6x1 = 78.
The equation indicates that 2x78 = 156g benzene will produce 6542kJ.
Using proportions you can then calculate that
x/6542kJ = 7.9g / 156g
x = 331.3kJ = 331300J.

heat = mass x ΔT x 4.18J/g°
ΔT = 331300J / (5691g x 4.18J/g°) = 13.9°

final temp = 21 + 14° = 35°C

4 0

4 years ago

Group the following electron configurations in pairs that would represent similar chemical properties at their atoms;

marishachu [46]

Answer: Pairs are: (a) and (d), (b) and (f), (c) and (e)

Explanation:

In a periodic table, elements are arranged in 18 vertical columns known as groups and 7 horizontal rows known as periods.

Elements arranged in a group show similar chemical properties because of the presence of same number of valence electrons.

Valence electrons are defined as the electrons which are present in the outermost shell of an atom. Outermost shell has the highest value of 'n' that is principal quantum number.

For the given options:

For a:

The given electronic configuration is: $1s^22s^22p^63s^2$

The number of valence electrons in the given configuration are 2

For b:

The given electronic configuration is: $1s^22s^22p^63s^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

For c:

The given electronic configuration is: $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For d:

The given electronic configuration is: $1s^22s^2$

The number of valence electrons in the given configuration are 2

For e:

The given electronic configuration is: $1s^22s^22p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For f:

The given electronic configuration is: $1s^22s^22p^63s^23p^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

Electronic configuration of (a) and (d) will form a pair, (b) and (f) will form a pair, (c) and (e) will form a pair and will have similar chemical properties.

Hence, the pairs are: (a) and (d), (b) and (f), (c) and (e)

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3 years ago

An astronaut weighs 104 newtons on the moon,where the strength of gravity is 1.6 newtons per kilogram

belka [17]

I'd say he ways about 35 kilograms, but I'm probably wrong, xD

6 0

4 years ago

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