Question

What volume of 0.08892 M HNO3 is required to react completetly with 0.2352 g of potassium hydrogen phosphate?

Answer 1

Answer:

0.0303 Liters

Explanation:

Given:

Mass of the potassium hydrogen phosphate = 0.2352

Molarity of the HNO₃ Solution = 0.08892 M

Now,

From the reaction it can be observed that 1 mol of potassium hydrogen phosphate reacts with 2 mol of HNO₃

The number of moles of 0.2352 g of potassium hydrogen phosphate

= Mass / Molar mass

also,

Molar mass of potassium hydrogen phosphate

= 2 × (39.09) + 1 + 30.97 + 4 × 16 = 174.15 g / mol

Number of moles = 0.2352 / 174.15 = 0.00135 moles

thus,

The number of moles of HNO₃ required for  0.00135 moles

= 2 ×  0.00135 mol of HNO₃

= 0.0027 mol of HNO₃

Now,

Molarity = Number of Moles / Volume

thus,

for 0.0027 mol of HNO₃, we have

0.08892 = 0.0027 / Volume

or

Volume =  0.0303 Liters

Answer 2

Answer: The volume of nitric acid is 0.025 L.

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of KHP = 0.2352 g

Molar mass of KHP = 204.22 g/mol

Putting values in above equation, we get:

$\text{Moles of KHP}=\frac{0.2352g}{204.22g/mol}=0.0011mol$

The chemical reaction for the formation of chromium oxide follows the equation:

$K_2HPO_4+2HNO_3\rightarrow H_2PO_4+KNO_3$

By Stoichiometry of the reaction:

1 mole of potassium hydrogen phosphate reacts with 2 mole of nitric acid

So, 0.0011 moles of potassium hydrogen phosphate will react with = $\frac{2}{1}\times 0.0011=0.0022mol$ of nitric acid

To calculate the volume of nitric acid, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

We are given:

Moles of Nitric acid = 0.0022 moles

Molarity of solution = 0.08892 M

Putting values in above equation, we get:

$0.08892mol/L=\frac{0.0022mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=0.025L$

Hence, the volume of nitric acid is 0.025 L.