Question

For the following reaction, 6.90 grams of tetraphosphorus decaoxide are mixed with excess perchloric acid (HClO4). The reaction yields 8.65 grams of phosphoric acid.

Answer 1

First, we have to write the equation:

`HClO_4\text{ + P}_4O_(10)\text{ }\rightarrow\text{ H}_3PO_4\text{ + Cl}_2O_7`

Then, we have to balance it:

After this, we have to make the respective calculations.

a. The first one is the theoretical yield. We have to calculate the molecular weight of each substance involved in the exercise:

`\begin{gathered} M.W.\text{ P}_4O_(10)\text{ = 31*4+16*10= 284 g/mol} \\ M.W.\text{ H}_3PO_4=\text{ 1*3+31+16*4= 98 g/mol} \end{gathered}`

Then, we calculate the produced quantity of phosphoric acid, in theory:

`6.90\text{ g P}_4O_(10)\text{ *}\frac{1\text{ mol P}_4O_(10)}{284\text{ g P}_4O_(10)}*\frac{4\text{ mol H}_3PO_4}{1\text{ mol P}_4O_(10)}*\frac{98\text{ g H}_3PO_4}{1\text{ mol H}_3PO_4}=\text{ 9.524 g H}_3PO_4`

Then, the answer is that the theoretical amount produced of phosphoric acid is 9.524 g.

b. Now, we have to apply the following formula:

`\%\text{ yield = }\frac{real\text{ produced amount}}{theoretical\text{ produced amount}}*100`

As we have those values, we replace them:

`\%yield=\text{ }\frac{8.65\text{ g}}{9.524\text{ g}}*100=\text{ 90.824\%}`

The answer is that the yield equals 90.824%