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2 years ago

3Fe+4H2O(yields) Fe3O4+4H2. What is the mole ratio of Fe3O4 to Fe?

Chemistry

2 answers:

WINSTONCH [101]2 years ago

7 0

Answer: The correct answer is Option A.

Explanation:

Mole ratio is defined as the ratio between the stoichiometric coefficients of the molecules present in the chemical reaction.

For the given balanced chemical equation:

$3Fe+4H_2O\rightarrow Fe_3O_4+4H_2$

By Stoichiometry of the reaction:

3 moles of iron metal reacts with 4 moles of water to produce 1 mole of iron oxide and 4 moles of hydrogen gas.

The mole ratio of $Fe_3O_4:Fe=1:3$

Hence, the correct answer is Option A.

RSB [31]2 years ago

3 0

Answer:

A- 1:3

Explanation:

Mole ratio is the ratio between the amount (In moles) of two substances involved in a chemical reaction.

For the reaction:

3Fe + 4H₂O → Fe₃O₄ + 4 H₂

There is 1 mole of Fe₃O₄ per 3 moles of Fe. In other words, the mole ratio of Fe₃O₄ to Fe is 1:3. Thus, answer is:

A- 1:3

I hope it helps!

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2 years ago

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3 years ago

A sample of sodium sulfite has a mass of 2.80 g.

Gnom [1K]

Answer:

For a: The number of sodium ions in given amount of sodium sulfite are $2.65\times 10^{22}$

For b: The number of sulfite ions in given amount of sodium sulfite are $1.325\times 10^{22}$

For c: The mass of one formula unit of sodium ions is $2.09\times 10^{-22}$ grams

Explanation:

The chemical formula of sodium sulfite is $Na_2SO_3$ . It is formed by the combination of 2 sodium $(Na^+)$ ions and 1 sulfite $(SO_3^{2-})$ ions

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of sodium sulfite = 2.80 g

Molar mass of sodium sulfite = 126 g/mol

Putting values in above equation, we get:

$\text{Moles of sodium sulfite}=\frac{2.80g}{126g/mol}=0.022mol$

For a:

Moles of sodium ions in sodium sulfite = (2 × 0.022) moles

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, 0.022 moles of sodium sulfite will contain = $(2\times 0.022\times 6.022\times 10^{23})=2.65\times 10^{22}$ number of sodium ions

Hence, the number of sodium ions in given amount of sodium sulfite are $2.65\times 10^{22}$

For b:

Moles of sulfite ions in sodium sulfite = (1 × 0.022) moles

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, 0.022 moles of sodium sulfite will contain = $(1\times 0.022\times 6.022\times 10^{23})=1.325\times 10^{22}$ number of sulfite ions

Hence, the number of sulfite ions in given amount of sodium sulfite are $1.325\times 10^{22}$

For c:

Molar mass of sodium sulfite = 126 g/mol

According to mole concept:

$6.022\times 10^{23}$ number of formula units are present in 1 mole of a compound

Or, $6.022\times 10^{23}$ number of formula units of sodium sulfite have a mass of 126 grams

So, 1 formula unit of sodium sulfite will have a mass of = $\frac{126}{6.022\times 10^{23}}\times 1=2.09\times 10^{-22}g$

Hence, the mass of one formula unit of sodium ions is $2.09\times 10^{-22}$ grams

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3 years ago