5.10 moles of H 2 (g) is mixed with 2.55 mol ofO 2 (g) and allowed to react as f

Question

5.10 moles of H2(g) is mixed with 2.55 mol ofO2(g) and allowed to react as follows:
2 H2(g) + O2(g) --> 2H2O(g)

How many atoms of H are initially present?
How many atoms of O are initially present?
How many atoms of H will there be in the product?
How many atoms of O will there be in the product?
How many moles of H2O will be formed if all theH2 and O2 react?
How many molecules of H2 were initially present?
How many molecules of O2 were initially present?
How many molecules of H2O were formed?

Explanation / Answer

First calculate which will be the limitant reactant. Supposeit is the oxygen. Then find how much hydrogen is consumed by the2.55 mol of oxygen. If it is less than 5.10 mol, the initial amountof hydrogen, the supposition is correct, otherwise it will be theoxygen. . 2 ---- 1 x -----2.55 x = 2.55*2/1 = 5.1 mol This means all reactives will be consumed, so:                   2H2(g) + O2(g) --> 2 H2O(g)
Initially         5.10         2.55            0 End                                                 5.10 . Since each mol contains 6.023x1023 molecules, nowyou can answer all questions using arithmetic. In themolecules where there are two atoms of the askes species, multiplyby two. How many atoms of H are initially present?
5.10( 6.023x1023)(2 atoms per molecule) =6.1434x1024
. How many atoms of O are initially present? 2.55(6.023x1023)(2) = 3.0717x1024 .
How many atoms of H will there be in the product?
5.10(6.023x1023)(2) = 6.1434x1024 . How many atoms of O will there be in the product?
5.10(6.023x1023) = 3.0717x1024 . How many moles of H2O will be formed if all theH2 and O2 react?
5.10(6.023x1023) = 3.0717x1024 . How many molecules of H2 were initiallypresent? 5.10(6.023x1023) = 3.0717x1024 .
How many molecules of O2 were initially present? 2.55(6.023x1023) = 1.5358x1024
. How many molecules of H2O were formed?
5.10(6.023x1023) = 3.0717x1024

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