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rosijanka [135]
3 years ago
5

When 2.0 mol h2o react with 2.0 mol fe in the reaction below, 1.5 mol of fe are used up in the reaction. how many moles of fe ar

e left over unreacted? 3 fe + 4 h2o → fe3o4 + 4h2?
Chemistry
2 answers:
GalinKa [24]3 years ago
7 0
The answer is 0.5 mol Fe left over.
Hope this helped! (:
Leni [432]3 years ago
7 0

Answer: 0.5 moles

Explanation: The given balanced equation is:

3Fe+4H_2O\rightarrow Fe_3O_4+4H_2

From the above equation, 4 moles of water reacts with 3 moles of Fe. So, mole ratio of water to Iron is 4:3.

If we take 2.0 moles of water and 2.0 moles of Fe then water will be the limiting reactant.

The given information says, 1.5 moles of Fe are used. We can also calculate how many moles of Fe are used as:

2.0molH_2O(\frac{3molFe}{4molH_2O})  = 1.5 mol Fe

So, 2.0 moles of Fe were taken and 1.5 moles of it were used then left over moles of Fe = 2.0 moles - 1.5 moles = 0.5 moles

Hence, 0.5 moles of Fe will be left over underacted.

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Dennis_Churaev [7]

Answer:

0.009725 moles of H2C2O4

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Explanation:

<u>Step 1</u>: The balanced equation

2MnO4- +5C2H2O4+6H+ →2Mn2+ +10CO2+8H2O

We can see that for 2 moles of Mno4- consumed , there is 5 moles of C2H2O4 needed and 6 moles H+ to produce 2 moles Mn2+, 10 moles of CO2 and 8 moles of H2O

<u>Step 2</u>: Calculate moles of MnO4-

Molarity = Moles/volume

Moles of Mno4- = Molarity of MnO4- * Volume of Mno4-

Moles of Mno4- = 0.1092M * 35.62 *10^-3 L

Moles of MnO4- = 0.00389 moles

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then for 0.00389 moles of MnO4-, there is 5/2 *0.00389 = <u>0.009725 moles of H2C2O4</u>

<u />

<u>Step 4:</u> Calculate moles of CaCO3

moles of H2C2O4 = moles CaCO3, therefore, 0.009725 moles H2C2O4 = 0.009725 moles CaCO3

<u>Step 5</u>: Calculate mass of CaCO3

Molar mass of CaCO3 = 100.09 g/mole

Mass of CaCO3 = moles of CaCO3 * Molar mass of CaCO3

Mass of CaCO3 = 0.009725 moles * 100.09 g/mole = 0.9734 g

<u>Step 6</u>: Calculate percentage by weight of CaCO3

Mass of CaCO3 = 0.9734g

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Mass percentage = 0.9734/1.2516 *100% = 77.77%

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Answer:

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