Question

All faculty members are happy to see students help each other. Dumbledore is particularly pleased with Hermione. Though, it should be mentioned that students should not simply copy off each other. You will not learn anything that way. Snape glares at Ron... Ron slouches in his chair. Snape thinks it’s time for a harder problem. How many milligrams of magnesium reacts with excess HCl to produce 31.2 mL of hydrogen gas at 754 Torr and 25.0◦C.The hydrogen is produced by the following reaction: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) Express your answer in milligrams.

Answer 1

Answer : The mass of
`Mg` required is 30.38 mg.

Explanation :

To calculate the moles of hydrogen gas, we use the equation given by ideal gas :

PV = nRT

where,

P = Pressure of hydrogen gas = 754 torr

V = Volume of the hydrogen gas = 31.2 mL = 0.0312 L

n = number of moles of hydrogen gas = ?

R = Gas constant =
`62.364\text{ L.torr }mol^(-1)K^(-1)`

T = Temperature of hydrogen gas =
`25^oC=273+25=298K`

Putting values in above equation, we get:

`754torr* 0.0312L=n* 62.364\text{ L.torr }mol^(-1)K^(-1)* 298K\\\\n=1.266* 10^(-3)mole`

Now we have to calculate the moles of
`Mg`.

The balanced chemical reaction is:

`Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)`

From the balanced chemical reaction, we conclude that

As, 1 mole of
`H_2` produced from 1 mole
`Mg`

As,
`1.266* 10^(-3)` mole of
`H_2` produced from
`1.266* 10^(-3)` mole
`Mg`

Now we have to calculate the mass of
`Mg`.

Molar mass of
`Mg` = 24 g/mol

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

`\text{Mass of }Mg=1.266* 10^(-3)mole* 24g/mole=30.38* 10^(-3)g=30.38mg`

conversion used : (1 g = 1000 mg)

Therefore, the mass of
`Mg` required is 30.38 mg