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2 years ago

How many grams of Fe2O3 are formed when 51.3 grams of iron, Fe, react completely with oxygen, O2?

Chemistry

1 answer:

lozanna [386]2 years ago

6 0

Answer: 73.4 g of $Fe_2O_3$ are formed when 51.3 grams of iron react completely with oxygen.

Explanation:

To calculate the moles :

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

$\text{Moles of} Fe=\frac{51.3g}{56g/mol}=0.92moles$

The balanced chemical reaction is:

$4Fe+3O_2\rightarrow 2Fe_2O_3$

As $Fe$ is the limiting reagent as it limits the formation of product.

According to stoichiometry :

4 moles of $Fe$ produce = 2 moles of $Fe_2O_3$

Thus 0.92 moles of $Fe$ will produce= $\frac{2}{4}\times 0.92=0.46moles$ of $Fe_2O_3$

Mass of $Fe_2O_3=moles\times {\text {Molar mass}}=0.46moles\times 159.69g/mol=73.4g$

Thus 73.4 g of $Fe_2O_3$ are formed when 51.3 grams of iron react completely with oxygen.

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Diatomic O2 can react with the element magnesium to form magnesium oxide (MgO). The balanced chemical equation is: 2Mg + O2 → 2M

tigry1 [53]

Answer:

We will expect 4 moles of MgO to be formed (option b).

Explanation:

Step 1: The balanced equation

2Mg + O2 → 2MgO

Step 2: Data given

Number of moles of Magnesium = 4 moles

Oxygen = in excess → this means Magnesium is the limiting reactant

Magnesium will completely be consumed ( 4 moles). There will remain 0 moles.

For 2 moles of magnesium consumed, we need 1 mole of oxygen to produce 2 moles of MgO.

For 4 moles of magnesium, we need 4/2 = 2 moles of oxygen.

For 4 moles of magnesium, we will produce 4/1 = 4 moles of MgO

We will expect 4 moles of MgO to be formed (option b).

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3 years ago

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2 years ago

What is the volume of 0.80 grams of o2 gas at stp? (5 points) group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 l

Vladimir [108]

Answer:

0.56L

Explanation:

This question requires the Ideal Gas Law: $PV=nRT$ where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.

Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units: $R=0.0821\frac{L\cdot atm}{mol\cdot K}$

Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means $T=273.15K$ and $P=1atm$

Lastly, we must calculate the number of moles of $O_2(g)$ there are. Given 0.80g of $O_2(g)$ , we will need to convert with the molar mass of $O_2(g)$ . Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2: $32g\text{ }O_2=1mol\text{ }O_2$