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Question 20 Two grams of magnesium are completely combusted with oxygen. The amount of oxygen (in grams) needed for this reaction is

A. 1.32 g.
B. 16.0 g.
C. 24.3 g.
D. 32.0 g

User Kaity
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1 Answer

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Final answer:

To find the grams of oxygen gas needed, we convert the given number of magnesium atoms to moles and use the stoichiometry of the reaction 2Mg + O2 → 2MgO to find the moles of O2 required, which is then converted to grams, resulting in 0.08 grams of O2 needed.

Step-by-step explanation:

To calculate the amount of oxygen gas needed to react with magnesium, we need to follow the stoichiometry of the balanced chemical equation for the combustion of magnesium:

2Mg + O2 → 2MgO

This indicates that 2 moles of magnesium react with 1 mole of oxygen gas to produce 2 moles of magnesium oxide.

To find the moles of magnesium, we use Avogadro's number (6.022 × 1023 atoms/mol). Thus, for 3.01 × 1021 atoms of magnesium:

Moles of Mg = (3.01 × 1021 atoms) / (6.022 × 1023 atoms/mol) = 0.005 moles of Mg
According to the balanced equation, we need half the number of moles of O2 as there are moles of Mg. Hence:

Moles of O2 = 0.005 moles of Mg / 2 = 0.0025 moles of O2

To convert moles of O2 to grams, we multiply by the molar mass of O2 (32 g/mol):

Mass of O2 = 0.0025 moles × 32 g/mol = 0.08 grams of O2

User Ebenezer
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