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Rom4ik [11]
3 years ago
14

When 26.62 miles of FeS2 reacts with 5.44 miles of O2 how many miles of SO2 are formed

Chemistry
1 answer:
Mademuasel [1]3 years ago
6 0
4 parts of FeS2 reacts with 11 parts of O2
so 26,62 moles of FeS2 would need 26,62/4 *11 = 73,2 moles of O2

So only 5,44 /11 *4 = 1,98 moles of FeS2 reacts with 5,44 moles of O2, O2 is the limiting
reactant
This means FeS2 is the reactant in excess and 26,62 - 5,44 = 21,18 moles of FeS2 are left over
and 1,98 * 8/4 = 3,96 moles of SO2 are formed
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Which factor plays the biggest role in delaying the detection of childhood<br> diseases?
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3 years ago
At 500 K in the presence of a copper surface , ethanoldecomposes according to the equation
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Answer:

Explanation:

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= - (250 -237 )/100 = - 13 / 100 torr/s

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How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl
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Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

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N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

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12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

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4 years ago
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