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Sagot :
Let's analyze the reaction:
[tex]\[ 4Na + O_2 \rightarrow 2Na_2O \][/tex]
Here, 4 moles of sodium (Na) react with oxygen (O[tex]\(_2\)[/tex]) to produce 2 moles of sodium oxide (Na[tex]\(_2\)[/tex]O). This stoichiometric ratio can be simplified to show that 2 moles of Na produce 1 mole of Na[tex]\(_2\)[/tex]O.
Given that we have 2.90 moles of Na, we need to calculate the corresponding moles of Na[tex]\(_2\)[/tex]O produced using the simplified ratio.
1. First, recognize the simplified stoichiometric ratio:
[tex]\[ 2 \, \text{moles of Na} \rightarrow 1 \, \text{mole of Na}_2O \][/tex]
2. Determine the moles of Na[tex]\(_2\)[/tex]O produced for 2.90 moles of Na:
[tex]\[ \text{Moles of Na\(_2\)O} = \frac{2.90 \, \text{moles of Na}}{2} = 1.45 \, \text{moles of Na}_2O \][/tex]
Therefore, there will be [tex]\(\boxed{1.45}\)[/tex] moles of Na[tex]\(_2\)[/tex]O produced.
[tex]\[ 4Na + O_2 \rightarrow 2Na_2O \][/tex]
Here, 4 moles of sodium (Na) react with oxygen (O[tex]\(_2\)[/tex]) to produce 2 moles of sodium oxide (Na[tex]\(_2\)[/tex]O). This stoichiometric ratio can be simplified to show that 2 moles of Na produce 1 mole of Na[tex]\(_2\)[/tex]O.
Given that we have 2.90 moles of Na, we need to calculate the corresponding moles of Na[tex]\(_2\)[/tex]O produced using the simplified ratio.
1. First, recognize the simplified stoichiometric ratio:
[tex]\[ 2 \, \text{moles of Na} \rightarrow 1 \, \text{mole of Na}_2O \][/tex]
2. Determine the moles of Na[tex]\(_2\)[/tex]O produced for 2.90 moles of Na:
[tex]\[ \text{Moles of Na\(_2\)O} = \frac{2.90 \, \text{moles of Na}}{2} = 1.45 \, \text{moles of Na}_2O \][/tex]
Therefore, there will be [tex]\(\boxed{1.45}\)[/tex] moles of Na[tex]\(_2\)[/tex]O produced.
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