a. how many grams of sodium oxide can be produced from the reaction between 5.74 g of na and excess o₂? (5…

Explanation:

Step1: Write the chemical equation

The reaction between sodium (Na) and oxygen ($O_2$) to form sodium oxide ($Na_2O$) is $4Na + O_2=2Na_2O$.

Step2: Calculate the molar - mass of Na

The molar - mass of Na ($M_{Na}$) is approximately $22.99\ g/mol$.

Step3: Calculate the number of moles of Na

The number of moles of Na ($n_{Na}$) is calculated using the formula $n=\frac{m}{M}$, where $m = 5.74\ g$ and $M = 22.99\ g/mol$. So, $n_{Na}=\frac{5.74\ g}{22.99\ g/mol}\approx0.25\ mol$.

Step4: Determine the mole ratio

From the balanced chemical equation, the mole ratio of $Na$ to $Na_2O$ is $4:2$ or $2:1$. So, if $n_{Na}=0.25\ mol$, then the number of moles of $Na_2O$ ($n_{Na_2O}$) is $n_{Na_2O}=\frac{0.25\ mol}{2}=0.125\ mol$.

Step5: Calculate the molar - mass of $Na_2O$

The molar - mass of $Na_2O$ ($M_{Na_2O}$) is $2\times22.99\ g/mol + 16.00\ g/mol=45.98\ g/mol+16.00\ g/mol = 61.98\ g/mol$.

Step6: Calculate the mass of $Na_2O$

The mass of $Na_2O$ ($m_{Na_2O}$) is calculated using the formula $m = n\times M$. So, $m_{Na_2O}=0.125\ mol\times61.98\ g/mol\approx7.75\ g$.

Answer:

$7.75\ g$