Chemistry, 23.11.2020 04:10 corcoranrobert1959

A sample of 5.0 g of hydrogen gas reacts with 5.0 g of oxygen gas. 2H2(g) + O2(g) > 2H2O

Please answer the questions in a b c d and e, or at least in any of the ones you know. Thank you.

A sample of 5.0 g of hydrogen gas reacts with 5.0 g of oxygen gas.

2H2(g) + O2(g) > 2H2O
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Answer from: coryoddoc3685

The percent yield of the reaction has been 74.47%.

The balanced chemical equation for the reaction has been:

$\rm 2\;H_2\;+\;O_2\;\rightarrow\;2\;H_2O$

(a) The theoretical yield can be calculated as:

The theoretical yield of water has been:

1 mole of oxygen produces = 2 moles of water

0.1562 moles of oxygen produces = 0.1562 $\times$ 2

= 0.312 moles of water.

The mass of water has been = moles $\times$ molecular weight

The theoretical yield of water = 0.312 $\times$ 18 grams

The theoretical yield of water = 5.616 grams.

(b) The limiting reactant has been the one whose concentration has been in a lesser quantity for the complete utilization of other reactants.

According to the reaction, for the complete utilization of 2 moles of Hydrogen, 1 mole of oxygen has been used.

The moles of 5 grams Hydrogen:

Moles = $\rm \dfrac{weight}{molecular\;weight}$

Moles of hydrogen = $\rm \dfrac{5}{1}$

Moles of hydrogen = 5 mol.

Moles of Oxygen = $\rm \dfrac{5}{32}$

Moles of oxygen = 0.1562 mol.

For the complete utilization of 5 moles of hydrogen 2.5 moles of oxygen has been required.

Since the oxygen has been present in a lesser quantity, the oxygen has been the limiting reactant.

(c) The consumption of 0.1562 moles of Oxygen has required 0.312 mol of Hydrogen. The hydrogen has been present in a higher quantity than required.

Thus. hydrogen has been the excess reactant.

(d) The remained excess reactant has been:

Remained = Total - consumed

Remained = 5 - 0.312

Remained = 4.68 mol.

The excess reactant has been 4.68 mol extra than required.

(e) The theoretical yield of water has been:

1 mole of oxygen produces = 2 moles of water

0.1562 moles of oxygen produces = 0.1562 $\times$ 2

= 0.312 moles of water.

The mass of water has been = moles $\times$ molecular weight

The theoretical yield of water = 0.312 $\times$ 18 grams

The theoretical yield of water = 5.616 grams.

The actual yield of water has been 4.2 grams.

The percent yield of water = $\rm \dfrac{Actual\;yield}{Theoretical\;yield}\;\times\;100$

The percent yield of water = $\rm \dfrac{4.2}{5.616}\;\times\;100$

The percent yield of water = 74.74 %

For more information about the limiting reactant, refer to the link:

link

Answer from: Laiyla27

a. 27%

b. H2 is the limiting reactant

c. O2 is the excess reactant

Answer from: Quest

43.452

step-by-step explanation:

284 times .153 = 43.452

Answer from: Quest

idk i’m struggling too

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A sample of 5.0 g of hydrogen gas reacts with 5.0 g of oxygen gas. 2H2(g) + O2(g) > 2H2O

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