What mass of H₂ is needed to react with 8.75 g of O₂ according to the following equation: O2(g) + H2(g) → H₂O(g)?

Chemistry

1 answer:

FromTheMoon [43]1 year ago

6 0

Explanation:

For reacting with 8.75 grams of oxygen, 1.08 grams of hydrogen is required.

The given balanced equation has been:

\rm O_2\;+\;2\;H_2\;\rightarrow\;H_2OO2+2H2→H2O

From the equation, 1 mole of oxygen reacts with 2 mole of hydrogen to give 1 mole of water.

The mass of oxygen has been: 8.75 g,

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

Moles of oxygen = \rm \dfrac{8.75}{32}328.75

Moles of oxygen = 0.27 mol

Since,

1 mole Oxygen = 2 mole hydrogen

0.21 mol oxygen = 0.54 mol hydrogen

Mass of hydrogen = moles \times× molecular weight

Mass of hydrogen = 0.54 \times× 2

Mass of hydrogen = 1.08 grams.

Thus, for reacting with 8.75 grams of oxygen, 1.08 grams of hydrogen is required.

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<h3>Reaction stoichiometry</h3>

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<h3>Moles of CO₂ formed</h3>

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<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

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