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3 years ago

Iodine chloride, ICl, can be made by the following reaction between iodine, I2, potassium iodate, KIO3, and hydrochloric acid.

2 answers:

6 0

The balanced chemical reaction is:

2 I2 + KIO3 + 6 HCl ---------> 5 ICl + KCl + 3 H2O

We are given the amount of the product to be produced from the reaction. This will be the starting point of our calculations.

28.6 g ICl (1 mol / 162.35 g ICl ) ( 2 mol I2 / 5 mol ICl ) ( 253.81 g I2 / 1 mol I2 ) = 17.88 g I2

svet-max [94.6K]3 years ago

5 0

Answer:

You need 17.87 grams to prepare 28.6 grams of ICl

Explanation:

In first place you should verify that the following chemical equation is balanced:

2 I₂ + KIO₃ + 6 HCl ⇒ 5 ICl + KCl + 3H₂O

For that, you must first look at the subscripts next to each atom to find the number of atoms the molecule has. For example, KIO₃ have 1 atom of K, 1 atom of I and 3 atoms of O.

Then you must pay attention to the coefficient in front of each molecule, which tells us the amount of moles of that molecule that react or form. For example, before I₂ there is a coefficient 2. This means that 2 moles of I₂ react. Then, to calculate the amount that react or are formed of each element, this coefficient is multiplied by the subindice of that element in each molecule. If the element appears in more than one molecule that reacts, the amounts are added. The same calculation is made with the products. So there are:

In reagents:

I: 5 atoms (coefficient 2 multiplied by subscript 2 of I₂ + coefficient 1 multiplied by subscript 1 of KIO₃)
K: 1 atom (coefficient 1 multiplied by subscript 1 of KIO₃)
O: 3 atoms (coefficient 1 multiplied by subscript 1 of KIO₃)
H: 6 atoms (coefficient 6 multiplied by subscript 1 of HCl)
Cl: 6 atoms (coefficient 6 multiplied by subscript 1 of HCl)

In products:

I: 5 atoms (coefficient 5 multiplied by subscript 1 of ICl)
K: 1 atom (coefficient 1 multiplied by subscript 1 of KCl)
O: 3 atoms (coefficient 3 multiplied by subscript 1 of H₂O)
H: 6 atoms (coefficient 3 multiplied by subscript 2 of H₂O)
Cl: 6 atoms (coefficient 5 multiplied by subscript 1 of ICl + coefficient 1 nultiplied by subscript 1 of KCl)

You can see that there is the same amount of each element in the reagents and in the products. So the chemical reaction is balanced.

Now you can calculate how many grams of iodine are needed to prepare 28.6 grams of ICl.

The relationships between the different molecules according to the chemical reaction are in moles, so you must first know when moles are 28.6 grams of ICl. For that you must know the molar mass of ICl, which is obtained with the molar mass of each element multiplied by the amount of that element in the molecule:

$massmolarof ICl=126.9 \frac{grams}{mol} +35.45 \frac{grams}{mol} = 162.35 \frac{grams}{mol}$

where 126.9 grams/mol is the mass of I and 35. 45 is the mass of Cl.

To determine how many moles are 28.6 grams you can do a simple rule of three:

If 162.35 grams are in 1 mole, how many moles are 28.6 grams?

$molesofICl=\frac{28.6grams*1mole}{162.35 grams}$

molesofICl=0.176 moles

Now you can use stoichiometric relationships, that is, the relationships that appear according to the chemical reaction, to calculate the amount of moles needed of I₂ to produce 0.176 moles of ICl. For that you do a simple rule of three:

If to produce 5 moles of ICl you need 2 moles of I₂ , how many moles of I₂ do you need to produce 0.176 moles of ICl?

$molesofI2=\frac{0.176molesofICl*2molesofI2}{5moles of ICl}$

molesofI2=0.0704 moles

So, you need 0.0704 moles of I₂ to produce 0.176 moles of ICl.

Now, knowing that the molar mass of I₂ is 253.8 grams (2*126.9 grams/mol), you can do a simple rule of three:

If 253.8 grams are in 1 mole, how many grams are in 0.0704 moles?

$gramsofI2=\frac{0.0704 moles*253.8grams}{1mole}$

gramsofI2=17.87 grams

This means that you need 17.87 grams to prepare 28.6 grams of ICl

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Answer:

a) the number-average molecular weight is 32,400 g/mol

b) the weight-average molecular weight is 36,320 g/mol

c) the degree of polymerization is 770

Explanation:

Given the data in the question;

Molecular Mean Number Weight Number Weight

weight average average

(g/mol) Mi(g/mol) xi wi xiMi wiMi

8,000-16,000 12,000 0.07 0.03 840 360

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∑ Mn=32,400 Mw=36,320

so;

a) the number-average molecular weight

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so from the table above; summation of Row Mixi

Mn = ∑Mixi = 32,400

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b) the weight-average molecular weight

Mw = ∑Miwi

so from the table above; summation of Row Miwi

Mw = ∑Miwi = 36,320

Therefore, the weight-average molecular weight is 36,320 g/mol

c) the degree of polymerization

the degree of polymerization of polypropylene can be determined using number-average molecular and repeat unit molecular weight.

now, for polypropylene { CH₂ = CH - CH₃ }

the repeat unit consist of 3 carbon atoms and 6 hydrogen atoms

given that;

Atomic weight of Carbon mC = 12.01 g/mol and

Atomic weight of Hydrogen mH = 1.008 g/mol

now we find the repeat unit molecular weight of polypropylene

m = nCmC + nHmH

where n is the number of repeat of atoms

so we substitute

m = ( 3 × 12.01) + ( 6 × 1.008)

m = 36.03 + 6.048

m = 42.078 g/mol

now we calculate the degree of polymerization;

DP = Mn / m

so we substitute

DP = 32,400 / 42.078

DP = 769.9985 ≈ 770

Therefore, the degree of polymerization is 770

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3.0 cm x 4.0 cm x 1.0 cm [?]cm^3

CaHeK987 [17]

Explanation:

Hi there!!

you asked to multiply these all right,

you can simply multiply it ,

=3cm × 4 cm × 1cm

= 12cm^2×1cm (4×3=12)

= 12cm^3 (12×1=12)

Therefore, theanswer is 12 cm^3.

Hope it helps..

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