Question

How many liters of a 3.0 M H3PO4 solution are required to react with 4.5 g of zinc? __ H3PO4 + __ Zn __ Zn3(PO4)2 + __ H2

Answer 1

Answer : The volume of $H_3PO_4$ is, 0.015 L

Solution :

First we have to calculate the moles of zinc.

$\text{Moles of zinc}=\frac{\text{Mass of zinc}}{\text{Molar mass of zinc}}=\frac{4.5g}{65g/mole}=0.069moles$

Now we have to calculate the moles of $H_3PO_4$.

The balanced reaction will be,

$2H_3PO_4+3Zn\rightarrow Zn_3(PO_4)_2+2H_2$

From the balanced reaction, we conclude that

As, 3 moles of zinc react with 2 moles of $H_3PO_4$

So, 0.069 moles of zinc react with $\frac{2}{3}\times 0.069=0.046$ moles of $H_3PO_4$

Now we have to calculate the volume of $H_3PO_4$

Formula used :

$Molarity=\frac{\text{Moles of }H_3PO_4}{\text{Volume of solution}}$

$3.0M=\frac{0.046mole}{\text{Volume of solution}}$

$\text{Volume of solution}=0.015L$

Therefore, the volume of $H_3PO_4$ is, 0.015 L

Answer 2

Answer:

Explanation:

The balanced chemical equation for the reaction is given as:

2 H₃PO₄ + 3 Zn ---> Zn₃(PO₄)₂  + 3 H₂

Given mass of Zn = 4.5 g

Molar mass of Zn = 65 g

Number of moles of Zn = Given mass/molar mass

                                       = 4.5 / 65

                                        = 0.069 moles

From the chemical equation:

3 moles of Zn requires 2 moles of H₃PO₄

⇒ 0.069 moles of Zn will require (2/3) x 0.069 moles of H₃PO₄

= 0.046 moles of H₃PO₄

Now we need to find how much liter of a 3.0 M H₃PO₄ will contain 0.046 moles of H₃PO₄.

Molarity is the number of moles of solute in 1 liter of solution.

A 3 M solution means there are 3 moles of H₃PO₄ in 1 liter of solution.

⇒ 3 moles of H₃PO₄  is present in 1 liter of solution.

⇒0.046 moles of H₃PO₄ will be present in 1/3 x 0.046 Liter of solution

⇒Volume of H₃PO₄  solution required is = 1/3 x 0.1 = 0.015 Liter.