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3 years ago

b) If 1.5 mole of oxygen reacted, how many mole of iron (III) oxide would be formed from the reaction?

Chemistry

1 answer:

lyudmila [28]3 years ago

7 0

Answer:

If there reacted 1.5 moles of O2, there will be produced 1.0 mol of Fe2O3

Explanation:

Step 1: Data given

Number of moles oxygen reacted = 1.5 moles

Step 2: The balanced equation

4Fe + 3O2 → 2Fe2O3

Step 3: Calculate moles of Fe2O3

For 4 moles Fe consumed, we need 3 moles of O2 to produce 2 moles of Fe2O3

For 1.5 moles O2 consumed, we'll have 2/3 * 1.5 = 1.0 mol of Fe2O3

If there reacted 1.5 moles of O2, there will be produced 1.0 mol of Fe2O3

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In a certain chemical reaction, 2 solid Mg atoms bond with O2 gas to produce solid

soldier1979 [14.2K]

Answer:

Option B. 2Mg(s) + O2 (g) —> 2MgO (s)

Explanation:

From the question given above,

We were told that:

2 solid Mg atoms bond with O2 gas to produce solid MgO.

This can be represented by an equation as follow:

2Mg(s) + O2 (g) —> MgO (s)

Next, we shall balance the above equation as follow:

2Mg(s) + O2 (g) —> MgO (s)

There are 2 atoms of Mg on the left side and 1 atom on the right side. It can be balance by putting 2 in front of MgO as shown below:

2Mg(s) + O2 (g) —> 2MgO (s)

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4 years ago

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Yes

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3 years ago

Two samples of the same compound are compared. what does the data represent? sample 1: 24.22 g carbon and 32.00 g oxygen sample

goblinko [34]

According to law of definite proportion:

In a compound, elements are always arranged in fixed ratio by mass.

Here, sample 1 has 23.22 g Carbon and 32.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

$n_{C}=\frac{m_{C}}{M_{C}}=\frac{24.22 g}{12 g/mol}\approx 2 mol$

Similarly, number of moles of oxygen will be:

$n_{O}=\frac{m_{O}}{M_{O}}=\frac{32 g}{16 g/mol}=2 mol$

The ratio of number of moles of carbon and oxygen will be:

$C:O=n_{C}:n_{O}=2:2=1:1$

Therefore, formula of compound will be CO.

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It has 36.22 g Carbon and 48.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

$n_{C}=\frac{m_{C}}{M_{C}}=\frac{36.22 g}{12 g/mol}\approx 3 mol$

Similarly, number of moles of oxygen will be:

$n_{O}=\frac{m_{O}}{M_{O}}=\frac{48 g}{16 g/mol}=3 mol$

The ratio of number of moles of carbon and oxygen will be:

$C:O=n_{C}:n_{O}=3:3=1:1$

The formula of compound will be CO.

Therefore, it is proved that carbon and oxygen are present in fixed ratios in both the samples.

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