Question

a scientist heated 10 grams of mercuric oxide and formed 9.3 grams of liquid mercury how many grams of oxygen were formed

Answer 1

Answer: 0.74 grams of oxygen were formed.

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}$  

For mercuric oxide:

Putting values in above equation, we get:

$\text{Moles of gold}=\frac{10g}{216.6g/mol}=0.046mol$

$\text{Moles of liquid mercury}=\frac{9.3g}{200.6g/mol}=0.046mol$

$2HgO\rightarrow 2Hg+O_2$

According to stochiometry,

2 moles of $HgO$ gives 1 mole of $O_2$

Thus 0.046 moles of $HgO$ gives=$\frac{1}{2}\times 0.046=0.023$ moles of $O_2$

Mass of $O_2$ produced=$moles\times {\text {Molar mass}}=0.023\times 32=0.74g$

Answer 2

Mercury oxide --- > liquid mercury plus oxygen.

If 10 grams of mercuric oxide were being heated, and 9.3 gram of liquid mercury is being produced. To find the grams of oxygen produced is, you take the mass of mercuric oxide minus the mass of liquid mercury which is 10 grams - 9.3 grams = 0.7 grams.

The answer is 0.7 grams.