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3 years ago

I need help finding the formulas of these ionic compounds

1 answer:

5 0

Li3PO4
Cs3N
KL
BaF2
BaO
Reaction of. An acid and a base so it consists of an cation and anion so then you take the name of the anion and then the name of the cation take the atomic symbols and depending on the group it’s in add the numbers ? I think x

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When using ion-selective electrodes, to compensate for a complex or unknown matrix, the _____________ method can be used to dete

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<h3>What are ion-selective electrodes?</h3>

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7 0

2 years ago

WILL MARK BRANILEST

ahrayia [7]

Answer:

FeCl3 is the limiting reactant

O2 is in excess

Theoretical yield Cl2 = 9.84 grams

The % yield is 96.5 %

Explanation:

Step 1: Data given

Mass of FeCl3 = 15.0 grams

Moles O2 = 4.0 moles

Mass of Cl2 produced = 9.5 grams

Step 2: The balanced equation

4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

Step 3: Calculate moles FeCl3

Moles FeCl3 = mass FeCl3 / molar mass FeCl3

Moles FeCl3 = 15.0 grams / 162.2 g/mol

Moles FeCl3 = 0.0925 moles

Step 4: Calculate limiting reactant

FeCl3 is the limiting reactant. Because we have way more (more than ratio 3:4) moles O2 than FeCl3. It will completely be consumed (0.0925 moles). O2 is in excess. There will react = 0.069375 moles O2

There will remain 4.0 - 0.069375 = 3.930625 moles

Step 5: Calculate moles Cl2

For 4 moles FeCl3 we need 3 moles O2 to produce 2 moles Fe2O3 and 6 moles Cl2

For 0.0925 moles FeCl3 moles we'll have 6/4 * 0.0925 = 0.13875 moles Cl2.

Step 6: Calculate mass Cl2

Mass Cl2 = moles * molar mass

Mass Cl2 = 0.13875 moles * 70.9 g/mol

Mass Cl2 = 9.84 grams

Step 7: Calculate % yield

% yield = (actual yield / theoretical yield) * 100%

% yield = (9.5 grams / 9.84 grams ) * 100%

% yield = 96.5 %

The % yield is 96.5 %

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4 years ago

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4 years ago

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How many atoms of oxygen are present in the reactants and products? 6CO2+6H2O-> C6H12O6+6O2

REY [17]

6 Carbon atoms
18 Oxygen atoms
12 Hydrogen atoms

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3 years ago