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

Answer:

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

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.