
If $48.6{\text{g}}$ of magnesium are placed in a container with 64 grams of oxygen gas and the reaction above proceeds to completion, what is the mass of ${\text{MgO}}\left( {\text{s}} \right)$ produced?
\[{\text{2Mg}}\left( {\text{s}} \right){\text{ + }}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to 2{\text{MgO}}\left( {\text{s}} \right)\]
A. $15.4$ grams
B. $32.0$ grams
C. $80.6$ grams
D. $96.3$ grams
E. $112$ grams
Answer
555k+ views
Hint: The above question can be solved on the basis of stoichiometry and mole concept. The reaction equation gives us the number of moles or molecules of reactants which react with each other to form products. We can find the mass of these reactants from their molar masses.
Formula: ${\text{mass}} = {\text{moles}} \times {\text{molar mass}}$
Complete Step by step Solution:
According to the given equation:
\[{\text{2Mg}}\left( {\text{s}} \right){\text{ + }}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to 2{\text{MgO}}\left( {\text{s}} \right)\],
Two moles of magnesium reacts with one mole of oxygen gas to form two moles of magnesium oxide.
The atomic weight of magnesium = 24 grams = 1 moles
Therefore 2 moles of magnesium = 48 grams.
Molecular weight of oxygen molecule = 32 grams
Molecular weight of magnesium oxide = 40 grams = 1 mole
Therefore 2 moles of magnesium oxide = 80 grams.
So, 48 grams of magnesium reacts with 32 grams of oxygen to form 80 grams of magnesium oxide.
Hence, 1 gram of magnesium reacts with $\dfrac{{32}}{{48}}$ grams of oxygen to form $\dfrac{{80}}{{48}}$ grams of magnesium oxide.
Hence, $48.6{\text{g}}$ of magnesium reacts with $\dfrac{{32}}{{48}} \times 48.6$ grams of oxygen to form $\dfrac{{80}}{{48}} \times 48.6$ grams of magnesium oxide = $80.6$ grams.
Hence, the correct answer is option C.
Notes: One mole of a substance is defined as the amount of substance present in $6.022 \times {10^{23}}$ $\text{molecules/ atoms/ other subatomic particles}$. This number is also called the Avogadro’s number or the Avogadro’s constant. The molecular weight of a compound and the atomic weight of an element contains one mole of the substance.
Formula: ${\text{mass}} = {\text{moles}} \times {\text{molar mass}}$
Complete Step by step Solution:
According to the given equation:
\[{\text{2Mg}}\left( {\text{s}} \right){\text{ + }}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to 2{\text{MgO}}\left( {\text{s}} \right)\],
| $Mg(s)$ | $O_2(g)$ | $MgO(s)$ |
| 2 moles | 1 mol | 2 moles |
Two moles of magnesium reacts with one mole of oxygen gas to form two moles of magnesium oxide.
The atomic weight of magnesium = 24 grams = 1 moles
Therefore 2 moles of magnesium = 48 grams.
Molecular weight of oxygen molecule = 32 grams
Molecular weight of magnesium oxide = 40 grams = 1 mole
Therefore 2 moles of magnesium oxide = 80 grams.
So, 48 grams of magnesium reacts with 32 grams of oxygen to form 80 grams of magnesium oxide.
Hence, 1 gram of magnesium reacts with $\dfrac{{32}}{{48}}$ grams of oxygen to form $\dfrac{{80}}{{48}}$ grams of magnesium oxide.
Hence, $48.6{\text{g}}$ of magnesium reacts with $\dfrac{{32}}{{48}} \times 48.6$ grams of oxygen to form $\dfrac{{80}}{{48}} \times 48.6$ grams of magnesium oxide = $80.6$ grams.
Hence, the correct answer is option C.
Notes: One mole of a substance is defined as the amount of substance present in $6.022 \times {10^{23}}$ $\text{molecules/ atoms/ other subatomic particles}$. This number is also called the Avogadro’s number or the Avogadro’s constant. The molecular weight of a compound and the atomic weight of an element contains one mole of the substance.
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