Question

A reaction container holds 5.77 g of $P_4$ and 5.77 g of $O_2$ and the following reaction occurs: $P_4+O_2\to P_4{O}_6.$ The amount of excess reagent left is?

• A. 10.51 g
• B. 2.84 g
• C. 20.32 g
• D. 1.31 g

Answer 1

The correct option is D 1.31 g

$P_4$ reacts with oxygen according to the equation:
$P_4+3O_2\to P_4{O}_6$
Moles of $O_2$ provided = $\frac{5.77}{32}=0.180$ Moles of $P_4$ provided = $\frac{5.77}{124}=0.046$
1 mole of $P_4$ reacts with 3 moles of Oxygen.
So, 0.046 moles of $P_4$ will react with $(3\times 0.046)=0.138$ moles of $O_2$
$P_4$ is the limiting reagent.
Therefore, $(4\times 31)$ g of $P_4$ reacts with $(3\times 32)$ g oxygen.
5.77 g will react with $\frac{(3\times 32)}{4\times 31}\times 5.77=4.46$ g of $O_2$
Amount of $O_2$ left $=5.77-4.46$ g = 1.31 g