Answer
$77.5 $ g of $Na_2O$ are produced.
Work Step by Step
1. Calculate the number of moles of $Na$:
Molar mass :
$Na: 22.99g$
22.99g = 22.99g
$57.5g \times \frac{1 mol}{ 22.99g} = 2.50mol (Na)$
- The balanced reaction is:
$4Na + O_2 --\gt 2Na_2O$
According to the coefficients, the ratio of $Na$ to $Na_2O$ is 4 to 2:
$2.5 mol (Na) \times \frac{ 2 mol(Na_2O)}{ 4 mol (Na)} = 1.25mol (Na_2O)$
2. Calculate the mass of $Na_2O$:
Molar mass :
$Na: 22.99g * 2= 45.98g $
$O: 16.00g$
45.98g + 16.00g = 61.98g
- Use the molar mass as a conversion factor:
$1.25 mol (Na_2O) \times \frac{ 61.98 g}{ 1 mol} = 77.5g (Na_2O)$
