Which of the following is a FALSE statement?

During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce

swat32

Answer: 16.32 g of $O_2$ as excess reagent are left.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol$

$\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol$

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

According to stoichiometry :

2 moles of $SO_2$ require = 1 mole of $O_2$

Thus 0.34 moles of $SO_2$ will require= $\frac{1}{2}\times 0.34=0.17moles$ of $O_2$

Thus $SO_2$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

Moles of $O_2$ left = (0.68-0.17) mol = 0.51 mol

Mass of $O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g$

Thus 16.32 g of $O_2$ as excess reagent are left.

3 0

2 years ago

Fake answers are reported and your account will be banned

Leona [35]

Answer:

BRAINLIEST?

Explanation:

Ammonia is a typical weak base. Ammonia itself obviously doesn't contain hydroxide ions, but it reacts with water to produce ammonium ions and hydroxide ions. My findings said that ammonia is a weak base, potassium hydroxide is a strong base, vinegar is a weak acid and ethyl alcohol is a weak acid.

Vinegar and ethyl alcohol are eliminated as they are acids. The question is on bases.... Potassium hydroxide is a strong base. So we are left with ammonia, being a weak base.

A is your answer

8 0

3 years ago

Read 2 more answers

Which metal can be extracted from bauxite ore?

Finger [1]

Aluminum
Bauxite ore is the world's primary source of aluminum. The ore must first be chemically processed to produce alumina (aluminum oxide). Alumina is then smelted using an electrolysis process to produce pure aluminum metal. Bauxite is typically found in topsoil located in various tropical and subtropical regions.

8 0

2 years ago

Read 2 more answers

Sodium hydroxide reacts with aluminum and water to produce hydrogen gas: 2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H

lianna [129]

Answer:

The mass of hydrogen gas formed is 0.205 grams

Explanation:

Step 1: Data given

Mass of 1.83 grams of Al

Mass of NaOH = 4.30 grams

Molar mass of Al = 26.98 g/mol

Molar mass of NaOH = 40 g/mol

Step 2: The balanced equation:

2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2(g)

Step 3: Calculate moles of Al

Moles Al = mass Al / Molar mass Al

Moles Al = 1.83 grams / 26.98 g/mol

Moles Al = 0.0678 moles

Step 4: Calculate moles of NaOH

Moles NaOH = 4.30 grams / 40 g/mol

Moles NaOH = 0.1075 moles

Step 5: Calculate limiting reactant

For 2 moles of Al, we need 2 moles of NaOH

Aluminium is the limiting reactant. It will completely be consumed ( 0.0678 moles)

NaOH is in excess. There will react 0.0678 moles

There will remain 0.1075 - 0.0678 = 0.0397 moles

Step 6: Calculate moles of hydrogen

For 2 moles of Al, we need 2 moles of NaOH, to produce 3 moles of hydrogen

For 0.0678 moles of Al, there is produced 0.0678 *3/2 = 0.1017 moles of H2

Step 7: Calculate mass of H2

Mass of H2 = Moles H2 * Molar mass of H2

Mass of H2 = 0.1017 moles * 2.02 g/mol

Mass of H2 = 0.205 grams

The mass of hydrogen gas formed is 0.205 grams

6 0

2 years ago

Give two examples of diffusion process in plants and animals.

lisabon 2012 [21]

answer:

in plants

Transport manufactured food from the leaves to others parts of the plant

Facilitates gaseous exchange through the stomata in the leaves to other parts of the plant

in animal

Exchange of respiratory gases across respiratory services

Excretion of nitrogenous waste in some unicellular organisms

Explanation:

Hope it benefit

3 0

2 years ago