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3 years ago

How many grams of chlorine gas must react to give 3.52g of BiCl3 according to the equation in exercise 23?

Chemistry

2 answers:

marshall27 [118]3 years ago

6 0

3.52g BiCl3 × 1 mol BiCl3/ 315.34g BiCl3 × 3 mol Cl/ 2 mol BiCl3 × 70.906g Cl/ 1 mol Cl= 1.187 g Cl

Brilliant_brown [7]3 years ago

6 0

The correct answer is 1.19 g of chlorine.

The following reaction is:

2Bi (s) + 3Cl₂ (g) ⇒ 2BiCl₃ (s)

In the reaction, it can be witnessed that 3 mol Cl₂ is equal to 2 mol BiCl₃

The molecular weight of BiCl₃ = 315.33

Thus,

3.52 g BiCl₃ = 3.52 g BiCl₃ × 1.00 mol BiCl₃ / 315.33 g BiCl₃

= 0.0112 mol BiCl₃

The mole ratio of Cl₂ and BiCl₃ is,

3 mol Cl₂ / 2 mol BiCl₃

Therefore, the amount of chlorine needed to form 0.0112 mol BiCl₃ is,

0.0112 mol BiCl₃ × 3 mol Cl₂ / 2 mol BiCl₃ = 0.0168 mol Cl₂

Now, the molecular weight of Cl₂ = 70.90

Thus,

0.0168 mol Cl₂ = 0.0168 mol Cl₂ × 70.90 g Cl₂ / 1.00 mol Cl₂

= 1.19 gm Cl₂

Hence, in the mentioned reaction, there is a need of 1.19 g of chlorine to react to produce 3.52 g of BiCl₃.

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larisa [96]

The answer is 200 J

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KE = 1/2 m * v²
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3 years ago

How many atoms are present in a sample of Potassium (K) weighing 33.49g?

madam [21]

Answer:

5.158 × 10²³ atoms K

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

33.49 g K

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of K - 39.10 g/mol

Step 3: Convert

$33.49 \ g \ K(\frac{1 \ mol \ K}{39.10 \ g \ K} )(\frac{6.022 \cdot 10^{23} \ atoms \ K}{1 \ mol \ K} )$ = 5.15797 × 10²³ atoms K

Step 4: Check

We are given 4 sig figs. Follow sig figs and round.

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3 years ago

An atom X has 2 electrons in its outer orbit. What will it do to form a stable ion?

matrenka [14]

Answer:

In a neutral molecule, the sum of the bonding valance electrons must be equal. So the products of the negative element and its charges and the positive element and its charge must be equal.

Explanation:

C1×N1 = C2×N2

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Balancing the two,

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2 years ago

For the following reaction, KcKc = 255 at 1000 KK.

bonufazy [111]

Answer :

The equilibrium concentration of CO is, 0.016 M

The equilibrium concentration of Cl₂ is, 0.034 M

The equilibrium concentration of COCl₂ is, 0.139 M

Explanation :

The given chemical reaction is:

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

Initial conc. 0.1550 0.173 0

At eqm. (0.1550-x) (0.173-x) x

As we are given:

$K_c=255$

The expression for equilibrium constant is:

$K_c=\frac{[COCl_2]}{[CO][Cl_2]}$

Now put all the given values in this expression, we get:

$255=\frac{(x)}{(0.1550-x)\times (0.173-x)}$

x = 0.139 and x = 0.193

We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.139

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The equilibrium concentration of Cl₂ = (0.173-x) = (0.173-0.139) = 0.034 M

The equilibrium concentration of COCl₂ = x = 0.139 M

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3 years ago