Question

Phosphine, an extremely poisonous and highly reactive gas, will react with oxygen to form tetraphosphorus decoxide and water, as described by the following UNBALANCED equation. PH3(g) + O2(g) → P4O10(s) + H2O(g) Calculate the mass of P4O10(s) formed when 225 g

Answer 1

Answer: The amount of
`P_4O_(10)` formed is 469.8 grams.

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` ......(1)

Given mass of phosphine = 225 g

Molar mass of phosphine = 34 g/mol

Putting values in equation 1, we get:

`\text{Moles of phosphine}=(225g)/(34g/mol)=6.62mol`

The given chemical reaction follows:

`4PH_3(g)+8O_2(g)\rightarrow P_4O_(10)(s)+6H_2O(g)`

Assuming that oxygen gas is present in excess, it is considered as an excess reagent.

Phosphine is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of phosphine produces 1 mole of
`P_4O_(10)`

So, 6.62 moles of phosphine will produce =
`(1)/(4)* 6.62=1.655mol` of
`P_4O_(10)`

Now, calculating the mass of
`P_4O_(10)` by using equation 1:

Molar mass of
`P_4O_(10)` = 283.9 g/mol

Moles of
`P_4O_(10)` = 1.655 moles

Putting values in equation 1, we get:

`1.655mol=\frac{\text{Mass of }P_4O_(10)}{283.9g/mol}\\\\\text{Mass of }P_4O_(10)=(1.655mol* 283.9g/mol)=469.8g`

Hence, the amount of
`P_4O_(10)` formed is 469.8 grams.