Question

What mass of phosphoric acid (H3PO4, 98.00 g/mol) is produced from the reaction of 10.00 g of P4O10 (283.89 g/mol) with 12.00 g water

Answer 1

The mass of phosphoric acid (H3PO4) produced from the reaction of 10.00 g of P4O10 with 12.00 g of water can be calculated through stoichiometry and is 13.8 g.

Step-by-step explanation:

Calculating the Mass of Phosphoric Acid

To calculate the mass of phosphoric acid (H3PO4) produced from the reaction between P4O10 and water (H2O), we'll need to consider the stoichiometry of the chemical reaction involved. The balanced chemical equation for this reaction is:

P4O10 + 6H2O → 4H3PO4

First, we convert the masses of P4O10 and H2O to moles using their molar masses, which are 283.89 g/mol (P4O10) and 18.015 g/mol (H2O) respectively. For P4O10:

• 
  
• 10.00 g P4O10 × (1 mol / 283.89 g) = 0.0352 mol of P4O10

and for H2O:

• 
  
• 12.00 g H2O × (1 mol / 18.015 g) = 0.666 mol of H2O

Next, we identify the limiting reagent by comparing the molar ratios from the balanced equation. There should be 6 moles of H2O for every mole of P4O10. Since we have more moles of H2O than needed according to this molar ratio, P4O10 is the limiting reagent.

Using the molar ratios from the equation, the moles of H3PO4 produced are 4 times the moles of P4O10:

• 
  
• 0.0352 mol P4O10 × (4 mol H3PO4 / 1 mol P4O10) = 0.1408 mol H3PO4

Finally, to find the mass of H3PO4 produced, we convert the moles of H3PO4 back to grams using its molar mass (98.00 g/mol):

• 
  
• 0.1408 mol H3PO4 × (98.00 g/mol) = 13.8 g H3PO4 produced

Answer 2

Answer: The mass of
`H_3PO_4` produced is, 13.82 grams.

Explanation : Given,

Mass of
`P_4O_(10)` = 10.00 g

Mass of
`H_2O` = 12.00 g

Molar mass of
`P_4O_(10)` = 283.89 g/mol

Molar mass of
`H_2O` = 18 g/mol

First we have to calculate the moles of
`P_4O_(10)` and
`H_2O`.

`\text{Moles of }P_4O_(10)=\frac{\text{Given mass }P_4O_(10)}{\text{Molar mass }P_4O_(10)}`

`\text{Moles of }P_4O_(10)=(10.0g)/(283.89g/mol)=0.0352mol`

and,

`\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}`

`\text{Moles of }H_2O=(12.0g)/(18g/mol)=0.666mol`

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

`P_4O_(10)+6H_2O\rightarrow 4H_3PO_4`

From the balanced reaction we conclude that

As, 1 mole of
`P_4O_(10)` react with 6 mole of
`H_2O`

So, 0.0352 moles of
`P_4O_(10)` react with
`0.0352* 6=0.211` moles of
`H_2O`

From this we conclude that,
`H_2O` is an excess reagent because the given moles are greater than the required moles and
`P_4O_(10)` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`H_3PO_4`

From the reaction, we conclude that

As, 1 mole of
`P_4O_(10)` react to give 4 mole of
`H_3PO_4`

So, 0.0352 mole of
`P_4O_(10)` react to give
`0.0352* 4=0.141` mole of
`H_3PO_4`

Now we have to calculate the mass of
`H_3PO_4`

`\text{ Mass of }H_3PO_4=\text{ Moles of }H_3PO_4* \text{ Molar mass of }H_3PO_4`

Molar mass of
`H_3PO_4` = 98.00 g/mole

`\text{ Mass of }H_3PO_4=(0.141moles)* (98.00g/mole)=13.82g`

Therefore, the mass of
`H_3PO_4` produced is, 13.82 grams.