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allochka39001 [22]
4 years ago
10

How many grams of C are required to react with 64.6 g of Fe2O3 ?

Chemistry
1 answer:
notsponge [240]4 years ago
4 0

Answer:

14.6 g

Explanation:

Fe₂O₃  +  3 C  →  2 Fe  +  3 CO₃

First, convert grams of Fe₂O₃ to moles.  The molar mass is 159.69 g/mol.

(64.6 g)/(159.69 g/mol) = 0.4045 mol

Now, use stoichiometry to convert moles of Fe₂O₃ to moles of C.  By looking at the chemical equation, you can see that for every 1 mole of Fe₂O₃, you need 3 moles of C.  Use this relationship.

(0.4045 mol Fe₂O₃) × (3 mol C/1 mol Fe₂O₃) = 1.214 mol C

Now, convert moles of C to grams.  The molar mass is 12.01 g/mol.

(1.214 mol C) × (12.01 g/mol) = 14.58 g

You need to use significant figures.

14.58 g ≈ 14.6 g

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Two thirds the number of moles of hydrogen

Explanation:

The balanced reaction is

3H₂ + N₂ → 2NH₃

So 1 mol of nitrogen gas reacts completely with 3 moles of hydrogen gas, to produce 2 moles of ammonia.

If equal moles of nitrogen and hydrogen are combined, <em>then hydrogen would be the limiting reactant</em>. Let's say we have 3 moles of each reactant, one mol of N₂ reacts with the 3 moles of H₂ and produces 2 moles of NH₃, and 2 moles of N₂ would remain. So the answer is<em> two thirds the number of moles of hydrogen.</em>

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Based on these values and on consideration of molecular geometry, the H-Se bond can be considered almost _____non-polar___ and the molecule is __polar_____.

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However, electro negativity difference alone is insufficient to determine the polarity of a molecule. The structure of the molecule is also considered. Based on the structure of the molecule, it is expected to have a dipole moment. Hence the molecule is polar.

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Which statement best describes the development of theories that connected microscopic and macroscopic phenomena?
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What are 7 elements from the periodic table that are named after objects in outer space?
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3 years ago
Determine the empirical formulas for compounds with the following percent compositions:
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Answer: a)  CS_2

b) CH_2O

Explanation:

a) If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 15.8g

Mass of S= 84.2 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{15.8g}{12g/mole}=1.32moles

Moles of S=\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{84.2g}{32g/mole}=2.63moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{1.32}{1.32}=1

For S =\frac{2.63}{1.32}=2

The ratio of C  : S = 1:2

Hence the empirical formula is CS_2

b) Mass of C= 40 g

Mass of H= 6.7 g

Mass of O = 53.3 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40g}{12g/mole}=3.33moles

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.7g}{1g/mole}=6.7moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.3g}{16g/mole}=3.33moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{3.33}{3.33}=1

For H = \frac{6.7}{3.33}=2

For O =\frac{3.33}{3.33}=1

The ratio of C : H: O= 1 :2: 1

Hence the empirical formula is CH_2O

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3 years ago
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