Question

fill blank boxes with the right answer. when 46 g of sodium is exposed to dry air, 62 g of sodium oxide is formed. what mass of oxygen was used? g o2

Explanation:

Step1: Apply mass - conservation principle

The mass of the product (sodium oxide) is the sum of the mass of sodium and the mass of oxygen used. Let the mass of oxygen be $m$. So, $m_{sodium\ oxide}=m_{sodium}+m_{oxygen}$.

Step2: Rearrange the formula to find mass of oxygen

We can rewrite the formula as $m_{oxygen}=m_{sodium\ oxide}-m_{sodium}$. Given $m_{sodium} = 46$ g and $m_{sodium\ oxide}=62$ g. Then $m = 62 - 46$.

Step3: Calculate the result

$m=16$ g.

Answer:

16