Question

Calculate the grams of oxygen in 90.0 g of cl2o

Answer 1

TO calculate the amount in mass of oxygen in the given compound we need data on the molar mass of the compound and oxygen. Also, we relate the number of moles of oxygen per 1 mol of the substance. We do as follows:

90.0 g ( 1 mol / 86.91 g ) ( 1mol O / 1 mol Cl2O) ( 16 g / 1 mol ) = 16.57 g O

Answer 2

Answer:

The correct answer is 16,576 g of $O_{2}$.

Explanation:

First, we must calculate the molar mass (Mm) of $O_{2}$ and $Cl_{2}O$.

Mm $O_{2}$= 32g

Mm $Cl_{2}O$ = 86,9 g

Now we know we have 90 g of $Cl_{2}O$, so let's find out how many moles of $Cl_{2}O$ there are in that amount.

90 g $Cl_{2}O$ * $\frac{1 mol Cl_{2}O}{86.9 g Cl_{2}O}$ = 1.036 mol $Cl_{2}O$

Now, in the $Cl_{2}O$ formation reaction we have that 1 mole of $O_{2}$ reacts to obtain 2 moles of $Cl_{2}O$:

$O_{2}$ + 2$Cl_{2}$ -> 2$Cl_{2}O$

Taking the coefficients, we calculate the number of moles of $O_{2}$ that react:

1,036 mol $Cl_{2}O$ * $\frac{1 mol O_{2}}{2 mol Cl_{2}O}$ = 0,518 mol $O_{2}$

Finally, we calculate the number of grams in the moles of $O_{2}$ that react:

0,518 mol $O_{2}$ * $\frac{32 g O_{2}}{1 mol O_{2}}$ = 16,576 g $O_{2}$

Have a nice day!