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2 years ago

Many grams of aluminum are required to produce 3.5 moles Al2O3 in the presence of excess O2?

1 answer:

4 0

The grams of aluminum that are required to produce 3.5 moles of AlO3 in presence of excess O2 is calculated as below

write the equation for reaction
4 Al + 3O2 =2 Al2O3

by use of mole ratio between Al to Al2O3 which is 4 :2 the moles of Al
=3.5 x4/2 = 7 moles

mass of Al = moles / x molar mass

= 7 moles x27 g/mol =189 grams

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Suppose the formation of iodine proceeds by the following mechanism:

NARA [144]

Answer:

Overall reaction

H2(g) + 2ICI(g) -----> I2(g) +2HCl(g)

Overall Rate = k1[H2] [ICl]

Explanation:

Overall reaction

H2(g) + 2ICI(g) -----> I2(g) +2HCl(g)

The overall reaction is the sum of the two two reactions shown in the question. After the two reactions are summed up properly, this overall reaction equation his obtained.

Since K1<<K2 it means that step 1 is slower than step 2. Recall that the rate if reaction depends on the slowest step of the reaction. Hence

Overall Rate = k1[H2] [ICl]

3 0

3 years ago

Calculate the mass of xenon difluoride gas with a volume of 0.223 L, pressure of 0.799 atm and temperature of 47.0 oC.

Greeley [361]

The mass of the gases can be determined by the moles of the gas in the ideal equation. The mass of xenon difluoride at 0.799 atm is 0.011 gms.

<h3>What is an ideal gas equation?</h3>

An ideal gas equation gives the moles of the substance from the temperature, volume, and pressure of the gas. The ideal gas equation can be shown as:

n = PV ÷ RT

Here, n = mass ÷ molar mass

Given,

Volume of xenon difluoride (V) = 0.223 L

Pressure of xenon difluoride (P) = 0.799 atm

Temperature of xenon difluoride (T) = 320.15 Kelvin

Gas constant (R) = 8.314 J⋅K⁻¹⋅mol⁻¹

The moles of the gas is calculated as:

n = PV ÷ RT

= 0.223 × 0.799 ÷ 8.314 × 320.15

= 0.1781 ÷ 2661.72

= 6.69 × 10⁻⁵ moles

Mass is calculated as:

169.29 × 6.69 × 10⁻⁵ = 0.011 gm

Therefore, 0.011 gms is the mass of xenon difluoride.

Learn more about ideal gas here:

brainly.com/question/14189621

#SPJ1

8 0

2 years ago

6. 7. A hyperbaric chamber has a volume of 200. L. (a) How many moles of oxygen are needed to fill the chamber at room temperatu

Otrada [13]

Answer:

a) 24.7 mol

b) 790 g

Explanation:

Step 1: Given data

Volume of the chamber (V): 200. L

Room temperature (T): 23 °C

Pressure of the gas (P): 3.00 atm

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 23°C + 273.15 = 296 K

Step 3: Calculate the moles (n) of oxygen

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 3.00 atm × 200. L/(0.0821 atm.L/mol.K) × 296 K = 24.7 mol

Step 4: Calculate the mass (m) corresponding to 24.7 moles of oxygen

The molar mass (M) of oxygen ga sis 32.00 g/mol. We will calculate the mass of oxygen using the following expression.

m = n × M

m = 24.7 mol × 32.00 g/mol = 790 g

6 0

3 years ago

How many atoms are in 1 mole of ca

Tju [1.3M]

Answer: 6.023 x 10^23

Explanation:

5 0

3 years ago

How many atoms of copper in 3. 11 g of copper

son4ous [18]

Atomic weight of copper = 63.5 g
1 mole of copper contains = 63.5g = 6.023 × 10²³ atoms of copper
1 g of copper contains
$= \frac{ 6.023 \times 10²³}{63.5}$ = 9.4 × 10²¹ atoms
Therefore, 3.11g of copper contains
$= 3.11 \times 9.4 \times {10}^{21} \\$
= 29.234 × 10²¹ atoms

Hope you could get an idea from here.

Doubt clarification - use comment section.

4 0

2 years ago