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Maksim231197 [3]
2 years ago
7

1. a) Anthony and Joseph create a start up company to make batteries for electric cars. Iron(III) phosphate is needed to create

the batteries. If 27.8 grams of iron (II) chloride reacts with 61.9 grams of sodium phosphate, what is the limiting reagent?
FeCl3(aq) + Na3PO4(aq) FePO4(s) + 3 NaCl(aq)
Chemistry
1 answer:
Elenna [48]2 years ago
3 0

Answer:

 LR is Na₃PO₄  

Explanation:

A quick way to determine the limiting reactant in a process is to convert reactant values to moles and then divide by the respective coefficient of the balanced equation. The smaller number of the division is the limiting reactant. For the given reaction, the rxn ratio of reactants is 1:1 so only the smaller mole value gives limiting reactant. However, if the reaction is NOT 1:1 the one must divide by respective coefficient to identify the smallest value and the limiting reactant.

This problem:

                          FeCl3(aq)          +    Na3PO4(aq)   => FePO4(s) + 3 NaCl(aq)

Given:                  27.8g                          61.9g

moles:        27.8g/162.2g/mole      1.9g/163.94g/mole

                    = 0.1714 mole               = 0.0116 mole

÷ coef.         => 0.1714/1 = 0.1714      => 0.0116/1 = 0.0116

smaller value is LR => =>  =>  =>  => => LR is Na₃PO₄              

Hope this helps. Doc :-)    

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Answer:

(a)57.48 percent (b) 29.45 percent

Explanation:

copper(II) bromide is 29.45 percent copper and 71.54 percent bromine. so the first element percentage composition is always the percentage composition of the compound.

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4 0
3 years ago
Which atom would tend to gain 1 valence electrons from another atom in order to become stable
Mrrafil [7]

Answer: Option (D) is the correct answer.

Explanation:

Atomic number of chlorine is 17 and its electronic distribution is 2, 8, 7.  

So, in order to attain stability it tends to gain one electron from a donor atom. Therefore, a chlorine ion is formed as Cl^{-1}.

As it gains only one electron that is why, -1 charge occurs.

Whereas Ar is a noble gas with atomic number 18. As it has completely filled octet so it will be unreactive in nature as it is already stable.

Similarly, He is also a noble gas so it is unreactive in nature.

Atomic number of boron (B) is 5 with electronic distribution 2, 3. So, in order to attain stability it needs to lose 3 electrons.

Thus, we can conclude that Cl is the atom which tends to gain 1 valence electron from another atom in order to become stable.

4 0
2 years ago
16 grams of propane, C 3 H 8 and 20 grams of oxygen, O 2 are reacted to produce carbon dioxide and water. Calculate the volume o
Veseljchak [2.6K]

Answer:

The volume of CO₂ produced is 8.4 L

Explanation:

The mass of propane in the reaction C₃H₈ = 20 grams

The mass of, oxygen, O₂ in the reaction = 20 grams

The produce of the reaction are carbon dioxide, CO₂ and water, H₂O

The balanced equation of the (combustion) reaction can be presented as follows;

C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)

Therefore, one mole of propane, C₃H₈, reacts with five moles of oxygen, O₂, to produce three moles of carbon dioxide, CO₂, and four moles of water molecules, H₂O, as steam

The number of moles = Mass/(Molar mass)

The molar mass of propane, C₃H₈ = 44.1 g/mol

The number of moles of propane in 16 grams of propane = 16/44.1 ≈ 0.3628 moles

The molar mass of oxygen, O₂ = 32.0 g/mol

The number of moles of oxygen in 20 grams of propane = 20/32 ≈ 0.625 moles

Therefore;

Given that 1 mole of C₃H₈ reacts with 5 moles of O₂

1 mole of O₂ will react with 1/5 moles of C₃H₈

0.625 moles of O₂ will react with 0.625/5 = 0.125 moles of C₃H₈ to produce 3 × 0.125 = 0.375 moles of CO₂

1 mole of an ideal gas occupies 22.4 L at standard temperature and pressure

Taking CO₂ as an ideal gas, we have;

0.375 mole of CO₂ will occupy 0.375 × 22.4 L = 8.4 L

Therefore, the volume of CO₂ produced = The volume occupied by the 0.375 moles of CO₂ = 8.4 L.

5 0
2 years ago
The concentration of a solution can be changed by _______ solute.
Snezhnost [94]
Add more solute..............
7 0
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