Which element is a metalloid? Li H Ge Ne

Methane and chlorine react to form four products: CH3Cl, CH2Cl2, CHCl3, and CCl4. At a particular temperature and pressure, 38.4

Anastasy [175]

Answer:

How many grams of CCL4 were formed? 116.9 g

How many grams of Cl2 reacted with the CH4? 243.8 g

Explanation:

First we need to know the molar mass for every element or compound in the reaction:

$M_{CH_{4}}=16 g/mol\\M_{CH_{3}Cl}=50.49g/mol\\M_{CH_{2}Cl_{2}}=84.93g/mol\\M_{CHCl_{3}}=119.38g/mol\\M_{CCl_{4}}=153.82g/mol$

Now we proceed to calculate the amount of moles produced, per product:

$n_{CH_{4}}=2.4\\n_{CH_{3}Cl}=0.18\\n_{CH_{2}Cl_{2}}=0.55\\n_{CHCl_{3}}=0.91\\n_{CCl_{4}}=n_{CH_{4}}-(n_{CH_{3}Cl}+n_{CH_{2}Cl_{2}}+n_{CHCl_{3}})\\n_{CCl_{4}}=0.76mol\\m_{CCl_{4}}=n_{CCl_{4}}*M_{CCl_{4}}\\m_{CCl_{4}}=116.9g$

To calculate the mass of chlorine we just need to make a mass balance:

$m_{CH_{4}}+m_{Cl_{2}}=m_{CH_{3}Cl}+m_{CH_{2}Cl_{2}}+m_{CHCl_{3}}+m_{CCl_{4}}\\m_{Cl_{2}}=m_{CH_{3}Cl}+m_{CH_{2}Cl_{2}}+m_{CHCl_{3}}+m_{CCl_{4}}-m_{CH_{4}}\\m_{Cl_{2}}=243.8g$

3 0

3 years ago

If an ideal gas at a constant temperature is initially at a pressure of 3.8 atm and is then allowed to expand to a volume of 5.6

BabaBlast [244]

Answer:

The most common example is the molar volume of a gas at STP (Standard Temperature and Pressure), which is equal to 22.4 L for 1 mole of any ideal gas at a temperature equal to 273.15 K and a pressure equal to 1.00 atm.If an ideal gas at a constant temperature is initially at a pressure of 3.8 atm and is then allowed to expand to a volume of 5.6 L and a pressure of 2.1 - 18914… ... of 5.6 L and a pressure of 2.1 atm, what is the initial volume of the gas? ... An ideal gas is at a pressure of 1.4 atm and has a volume of 3 L.

Explanation:

I hope I help :)

7 0

3 years ago

What type of bonds are located between the nucleotides found in opposing DNA strands (i.e. the strands that are across from each

belka [17]

Answer:

hydrogen bond are found between nucleotides

7 0

2 years ago

Consider the group a1 element sodium atomic number 11, the group 3a element aluminum atomic number 13, and the group 7a element

Akimi4 [234]

According to there chemical properties

7 0

4 years ago

PCl3(g) + Cl2(g) ⇋ PCl5(g) Kc = 91.0 at 400 K. What is the [Cl2] at equilibrium if the initial concentrations were 0.24 M for PC

Dmitry_Shevchenko [17]

Answer:

[Cl₂] in equilibrium is 1.26 M

Explanation:

This is the equilibrium:

PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)

Kc = 91

So let's analyse, all the process:

PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)

Initially 0.24 M 1.50M 0.12 M

React x x x

Some amount of compound has reacted during the process.

In equilibrium we have

0.24 - x 1.50 - x 0.12 + x

As initially we have moles of product, in equilibrium we have to sum them.

Let's make the expression for Kc

Kc = [PCl₅] / [Cl₂] . [PCl₃]

91 = (0.12 + x) / (0.24 - x) ( 1.50 - x)

91 = (0.12 + x) / (0.36 - 0.24x - 1.5x + x²)

91 (0.36 - 0.24x - 1.5x + x²) = (0.12 + x)

32.76 - 158.34x + 91x² = 0.12 +x

32.64 - 159.34x + 91x² = 0

This a quadratic function:

a = 91; b= -159.34; c = 32.64

(-b +- √(b² - 4ac)) / 2a

Solution 1 = 1.5

Solution 2 = 0.23 (This is our value)

So [Cl₂] in equilibrium is 1.50 - 0.23 = 1.26 M

5 0

4 years ago