
How do I convert grams to moles to mole ratio ?
\[Mg\; + {\text{ }}{O_2}^{}\;\; - - - > \;MgO\;16.2g\; + 10.67g\; = {\text{ }}26.86g\]
Answer
444k+ views
Hint: The empirical formula is created, the molecular formula for a compound can be resolute if the molar mass of the compound is identified. Basically calculate the mass of the empirical formula and divide the molar mass of the compound by the mass of the empirical formula to discover the ratio between the molecular formula and the empirical formula. Multiply all the atoms by this ratio to catch the molecular formula.
Complete step by step answer:
For appetizers, make assured that the chemical equation is balanced
\[2M{g_{(s)}} + {\text{ }}{O_{2(g)}} \to \;2Mg{O_{(s)}}\]
Sign that the response consumes \[2\] moles of magnesium and \[1\] mole of oxygen gas and produces \[2\] moles of magnesium oxide.
You can hence say that the number of moles of each chemical type that takes part in this reaction must gratify.
Currently, you can change grams to moles by using molar mass. In this case, you have
\[\begin{array}{*{20}{l}}
{{M_M}Mg = 24.3050{\text{ }}g{\text{ }}mo{l^{ - 1}}} \\
{{M_M}O2 = 32.0{\text{ }}g{\text{ }}mo{l^{ - 1}}} \\
{{M_M}MgO = 40.3050{\text{ }}g{\text{ }}mo{l^{ - 1}}}
\end{array}\]
So, change the mass of magnesium from grams to simply, use the element's molar mass as a renovation factor.
\[16.2g*{\text{ }}(1{\text{ }}mole{\text{ }}Mg{\text{ }}/{\text{ }}24.3050g)\]= \[0.6665\;\] moles Mg
Do the same for the example of oxygen gas.
\[10.67g*{\text{ }}(1{\text{ }}mole{\text{ }}{O_2}/{\text{ }}32.0g)\]= \[0.3334\;\] moles $O_2$
To conclude, do the similar for the magnesium oxide.
\[26.86g*{\text{ }}(1{\text{ }}mole{\text{ }}MgO{\text{ }}/40.3050g)\]= \[0.6664\;\] moles MgO
Thus , you see that when \[0.6665\;\] moles of magnesium react with \[0.3334\;\] moles of oxygen gas, you get \[0.6664\;\] moles of magnesium oxide.
The principles are not a strict match because of turning and of the values used for the molar masses of the three chemical types.
Note:
Stoichiometry agrees to make calculations about the outcomes of chemical reactions.
-Calculate the mass of a product of a chemical reaction if given the starting masses of reactants.
-Calculate the volume of a gas which will be produced by a reaction if given the first amounts of reactants.
-Define the optimum ratio of reactants for a chemical reaction so that all reactants are fully used.
Complete step by step answer:
For appetizers, make assured that the chemical equation is balanced
\[2M{g_{(s)}} + {\text{ }}{O_{2(g)}} \to \;2Mg{O_{(s)}}\]
Sign that the response consumes \[2\] moles of magnesium and \[1\] mole of oxygen gas and produces \[2\] moles of magnesium oxide.
You can hence say that the number of moles of each chemical type that takes part in this reaction must gratify.
moles Mg | moles $O_2$ | moles MgO |
2 | 1 | 2 |
Currently, you can change grams to moles by using molar mass. In this case, you have
\[\begin{array}{*{20}{l}}
{{M_M}Mg = 24.3050{\text{ }}g{\text{ }}mo{l^{ - 1}}} \\
{{M_M}O2 = 32.0{\text{ }}g{\text{ }}mo{l^{ - 1}}} \\
{{M_M}MgO = 40.3050{\text{ }}g{\text{ }}mo{l^{ - 1}}}
\end{array}\]
So, change the mass of magnesium from grams to simply, use the element's molar mass as a renovation factor.
\[16.2g*{\text{ }}(1{\text{ }}mole{\text{ }}Mg{\text{ }}/{\text{ }}24.3050g)\]= \[0.6665\;\] moles Mg
Do the same for the example of oxygen gas.
\[10.67g*{\text{ }}(1{\text{ }}mole{\text{ }}{O_2}/{\text{ }}32.0g)\]= \[0.3334\;\] moles $O_2$
To conclude, do the similar for the magnesium oxide.
\[26.86g*{\text{ }}(1{\text{ }}mole{\text{ }}MgO{\text{ }}/40.3050g)\]= \[0.6664\;\] moles MgO
Thus , you see that when \[0.6665\;\] moles of magnesium react with \[0.3334\;\] moles of oxygen gas, you get \[0.6664\;\] moles of magnesium oxide.
The principles are not a strict match because of turning and of the values used for the molar masses of the three chemical types.
Note:
Stoichiometry agrees to make calculations about the outcomes of chemical reactions.
-Calculate the mass of a product of a chemical reaction if given the starting masses of reactants.
-Calculate the volume of a gas which will be produced by a reaction if given the first amounts of reactants.
-Define the optimum ratio of reactants for a chemical reaction so that all reactants are fully used.
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