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3 years ago

Iron forms both 2+ and 3+ cations. Write formulas for the oxide, sulfide, and chloride compounds of the 3+ iron cation.

Chemistry

1 answer:

nordsb [41]3 years ago

4 0

Answer:

Fe2O3, Fe2S3 and FeCl3

Explanation:

The formular shows the number of atoms of element present in the compound. The number of atoms in turn depends on the charges. The goal is to have a neutral compound with oxidation number of zero.

Oxide:

Iron(Fe) has oxidation number (O.N) of +3, oxygen has O.N of -2. The formular with a balanced (Charge = 0) O.N number is; Fe2O3

Sulphide:

Iron(Fe) has oxidation number (O.N) of +3, sulphur has O.N of -2. The formular with a balanced (Charge = 0) O.N number is; Fe2S3

Chloride:

Iron(Fe) has oxidation number (O.N) of +3, chlorine has O.N of -1. The formular with a balanced (Charge = 0) O.N number is; FeCl3

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Fe2O3(s) + 3CO(g) ---> 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast

baherus [9]

Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%

Explanation:

To calculate the moles :

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

$\text{Moles of} Fe_2O_3=\frac{450}{160}=2.8moles$

$\text{Moles of} CO=\frac{260}{28}=9.3moles$

$Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)$

According to stoichiometry :

1 mole of $Fe_2O_3$ require 3 moles of $CO$

Thus 2.8 moles of $Fe_2O_3$ will require= $\frac{3}{1}\times 2.8=8.4moles$ of $CO$

Thus $Fe_2O_3$ is the limiting reagent as it limits the formation of product and $CO$ is the excess reagent.

As 1 mole of $Fe_2O_3$ give = 2 moles of $Fe$

Thus 2.8 moles of $Fe_2O_3$ give = $\frac{2}{1}\times 2.8=5.6moles$ of $Fe$

Mass of $Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g$

Theoretical yield of liquid iron = 313.6 g

Experimental yield = 288 g

Now we have to calculate the percent yield

$\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{288g}{313.6g}\times 100=91.8\%$

Therefore, the percent yield is, 91.8%

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The reaction for photosynthesis producing glucose sugar and oxygen gas is:

Anvisha [2.4K]

1.5 grams of glucose is produced from 2.20 g of CO₂.

To find the mass of glucose produced, first you must know the balanced reaction. For this, the Law of Conservation of Matter is followed.

The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

So, in this case, the balanced reaction is:

6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the amounts of moles of each reactant and product participate in the reaction:

CO₂: 6 moles
H₂O: 6 moles
C₆H₁₂O₆: 1 mole
O₂: 6 moles

So, you know that 2.20 g of CO₂ react, whose molar weight is 44.01 g/mole. By definition of molar mass, 1 mole of CO₂ has 44.01 g. So, the number of moles that 2.20 grams of the compound represent is calculated as:

$moles of CO_{2} =2.20 grams*\frac{1 mole}{44.01 grams}$

moles of CO₂= 0.05 moles

Now you must follow the following rule of three: if by stoichiometry of the reaction 6 moles of CO₂ produce 1 mole of C₆H₁₂O₆, 0.05 moles of CO₂ produce how many moles of C₆H₁₂O₆?

$moles of C_{6} H_{12} O_{6} =\frac{0.05moles of CO_{2} *1 mole of C_{6} H_{12} O_{6}}{6moles of CO_{2}}$

moles of C₆H₁₂O₆= 8.33*10⁻³

Being the molar mass of glucose 180.18 g/mole, the mass that 8.33*10⁻³ moles of the compound represent is calculated as:

$mass of glucose =8.33*10^{-3} moles*\frac{180.18 grams}{1 mole}$

mass of glucose= 1.5 grams

Then, 1.5 grams of glucose is produced from 2.20 g of CO₂.

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