Q. 7.5 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate each of the following quantities in $2.00$ moles of ${\mathrm{H}}_3{\mathrm{PO}}_4$ :

1. moles of $\mathrm{H}$
   
2. moles of $\mathrm{O}$
   
3. atoms of $\mathrm{P}$
4. atoms of  $\mathrm{O}$
   

Step-by-Step Solution

Answer

Quantities present in $2.00$ moles of ${\mathrm{H}}_3{\mathrm{PO}}_4$ are:

1. $6.00$ moles of hydrogen 
2. $8.00$ moles of oxygen
3. $12.04\times {10}^{23}$ atoms of phosphorus
4. $48.16\times {10}^{23}$ atoms of oxygen

Part (a) Step 1: Given Information

We need to find number of moles of $\mathrm{H}$ in $2.00$ moles of ${\mathrm{H}}_3{\mathrm{PO}}_4$

Part(a) Step 2: Explanation

One mole of ${\mathrm{H}}_3{\mathrm{PO}}_4$ contains $3.00$ moles of $\mathrm{H}$

So, moles of $\mathrm{H}$ in two moles of ${\mathrm{H}}_3{\mathrm{PO}}_4$ = $2\times 3=6.00$

Hence, $6.00$moles of hydrogen present in $2.00$ moles of ${\mathrm{H}}_3{\mathrm{PO}}_4$.

Part (b) Step 1: Given Information

We need to find number of moles of $\mathrm{O}$ in $2.00$ moles of  ${\mathrm{H}}_3{\mathrm{PO}}_4$

Part(b) Step 2: Explanation

One mole of ${\mathrm{H}}_3{\mathrm{PO}}_4$ contains $4.00$ moles of  $\mathrm{O}$

So, moles of $\mathrm{O}$ in two moles of ${\mathrm{H}}_3{\mathrm{PO}}_4$ = $2\times 4=8.00$

Hence,$8$ moles of oxygen is  present in $2$ moles of ${\mathrm{H}}_3{\mathrm{PO}}_4$.

Part (c) Step 1: Given Information

We need to find number of  atoms of $\mathrm{P}$ in $2.00$ moles of ${\mathrm{H}}_3{\mathrm{PO}}_4$

Part(c) Step 2: Explanation

One mole of  ${\mathrm{H}}_3{\mathrm{PO}}_4$ contains one mole of $\mathrm{P}$ in it.

Number of atoms of $\mathrm{P}$ in $2.00$ moles of ${\mathrm{H}}_3{\mathrm{PO}}_4$= $2\times 6.02\times {10}^{23}=12.04\times {10}^{23}$

Hence, $12.04\times {10}^{23}$ atoms are present.

Part (d) Step 1: Given Information

We need to find number of  atoms of $\mathrm{O}$ in $2.00$ moles of ${\mathrm{H}}_3{\mathrm{PO}}_4$

Part(d) Step 2: Explanation

One mole of  ${\mathrm{H}}_3{\mathrm{PO}}_4$ contains four mole of $\mathrm{O}$ in it.

Number of atoms of oxygen in two moles of ${\mathrm{H}}_3{\mathrm{PO}}_4$ = $$\begin{gathered} 2\times 4\times 6.02\times {10}^{23} \\ =48.16\times {10}^{23} \end{gathered}$$

Hence, $48.16\times {10}^{23}$ atoms are present.