Question

Magnesium metal (0.100 mol) and a volume of aqueous hydrochloric acid that contains 0.500 mol of HCl are combined and react to completion. How many liters of hydrogen gas, measured at STP, are produced? Mg(s) + 2HCl(aq) → MgCl2 (aq) + H2 (g)

Answer 1

2.24 L of hydrogen gas, measured at STP, are produced.

Step-by-step explanation:

Given, Moles of magnesium metal,
`Mg` = 0.100 mol

Moles of hydrochloric acid,
`HCl` = 0.500 mol

According to the reaction shown below:-

`Mg_((s)) + 2HCl_((aq))\rightarrow MgCl_2_((aq)) + H_2_((g))`

1 mole of Mg reacts with 2 moles of
`HCl`

0.100 mol of Mg reacts with 2*0.100 mol of
`HCl`

Moles of
`HCl` must react = 0.200 mol

Available moles of
`HCl` = 0.500 moles

Limiting reagent is the one which is present in small amount. Thus,
`Mg` is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of
`Mg` on reaction forms 1 mole of
`H_2`

0.100 mole of
`Mg` on reaction forms 0.100 mole of
`H_2`

Mole of
`H_2` = 0.100 mol

At STP,

Pressure = 1 atm

Temperature = 273.15 K

Volume = ?

Using ideal gas equation as:

`PV=nRT`

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V L = 0.100 × 0.0821 L.atm/K.mol × 273.15 K

⇒V = 2.24 L

2.24 L of hydrogen gas, measured at STP, are produced.