Select all that apply.

If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.

kiruha [24]

Answer:

Sr(OH)₂ will be the limiting reagent.

Explanation:

First of all, you should know the following balanced chemical equation:

2 H₂CO₃ + 2 Sr(OH)₂ → 4 H₂O + Sr₂(CO₃)₂

The balanced equation is based on the Law of Conservation of Mass, which says that matter cannot be created or destroyed. Therefore, the number of each type of atom on each side of a chemical equation must be the same.

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). By stoichiometry the following amounts in moles react:

strontium hydroxide: 2 moles
carbonic acid: 2 moles

Now, you know the following masses of the elements:

Sr: 87.62 g/mole
O: 16 g/mole
H: 1 g/mole

So the molar mass of strontium hydroxide is:

Sr(OH)₂= 87.62 g/mole + 2*(16 g/mole + 1 g/mole)= 121.62 g/mole

You apply the following rule of three, if 121.62 grams of hydroxide are present in 1 mole, 12.5 grams in how many moles are they?

$moles of strontium hydroxide=\frac{12.5 grams*1 mole}{121.62 grams}$

moles of hydroxide= 0.103 moles

On the other hand, you have 150 ml of 3.5 M carbonic acid. Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you can apply the following rule of three: if in 1 L there are 3.5 moles of carbonic acid, in 0.150 L (being 1 L = 1000 mL, 0.150 L = 150 mL) how many moles of acid are there?

$molesofcarbonicacid=\frac{0.150 L*3.5 moles}{1 L}$

moles of carbonic acid= 0.525 moles

Finally, to calculate the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 2 mole of strontium hydroxide reacts with , how much moles of carbonic acid will be needed if 0.103 moles of strontium hydroxide react?

$molesofcarbonicacid=\frac{0.103 moles of strontium hydroxide*2 moles of carbonic acid}{2 moles of strontium hydroxide}$

moles of carbonic acid= 0.103 moles

But 0.525 moles are available. Since more moles are available than you need to react with 0.103 moles of strontium hydroxide, Sr(OH)₂ will be the limiting reagent.

7 0

3 years ago

O que é VAPORIZAÇÃO, CONDENSAÇÃO, SUBLIMAÇÃO, FUSÃO, SOLIDIFICAÇÃO

ipn [44]

Answer:

Fusão: a substância volta do sólido para o líquido. Condensação: a substância muda de um gás para um líquido. Vaporização: a substância muda de líquido para gás. Sublimação: a substância muda diretamente de um sólido para um gás sem passar pela fase líquida.

7 0

3 years ago

Calculate the number of grams of oxygen gas in a 24L container at STP

romanna [79]

There are 34 g of oxygen in the container.

We can use the Ideal Gas Law to solve this problem.

$pV = nRT$

But $n = \frac{m}{M}$ , so

$pV = \frac{m}{M}RT$ and

$m = \frac{pVM}{RT}\\$

STP is 0 °C and 1 bar, so

$m = \frac{\text{1 bar} \times \text{24 L} \times 32.00 \text{ g}\cdot\text{mol}^{-1}}{\text{0.083 14 } \text{bar}\cdot\text{L}\cdot\text{K}^{-1}\text{mol}^{-1}\times\text{273.15 K} } = \textbf{34 g}\\$

7 0

3 years ago

There is a 50 g sample of ra-229. It has a half-life of 4 minutes.how much will be left after 12 minutes? a. 3.13 g b. 6.25 g c.

polet [3.4K]

It's B. 6.25

first divide 50/2

then divide 25/2

after that divide 12.5/2

Then you get 6.25

Hope it helps!

3 0

3 years ago

If 156.06 g of propane, C3H8, is burned in excess oxygen, how many grams of water are formed? C3H8 + O2 → CO2 + H2O Select one:

Nastasia [14]

Answer:

The correct option is;

a. 255.0 g

Explanation:

The given information are;

Mass of propane, C₃H₈ in the combustion reaction = 156.06 g

The equation of the combustion reaction is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

From the balanced chemical equation of the reaction, we have;

One mole of propane, C₃H₈ reacts with five moles oxygen gas, O₂, to form three moles of carbon dioxide, CO₂, and four moles of water, H₂O

The molar mass of propane gas = 44.1 g/mol

The number of moles, n, of propane gas = Mass of propane gas/(Molar mass of propane gas) = 156.06/44.1 = 3.54 moles

Given that one mole of propane gas produces 4 moles of water molecule (steam) H₂O, 3.54 moles of propane gas will produce 4×3.54 = 14.16 moles of (steam) H₂O

The mass of one mole of H₂O = 18.01528 g/mol

The mass of 14.16 moles of H₂O = 14.16 × 18.01528 = 255.0 g

The mass of H₂O produced = 255.0 g

3 0

3 years ago