Question

Magnesium metal (0.100 mol) and hydrochloric acid (0.500 mol HCl) are combined and react to completion. What volume of hydrogen gas, measured at STP, is produced? Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) (R = 0.08206 L • atm/K • mol) Select one: A. 22.4 L of H2 B. 5.60 L of H2 C. 4.48 L of H2 D. 11.2 L of H2 E. 2.24 L of H2

Answer 1

A.
`V=2.24L` of
`H_(2)`

Step-by-step explanation:

1. First take the balanced chemical equation to obtain hydrogen gas:

`Mg(s)+2HCl(aq)=MgCl_(2)(aq)+H_(2)(g)`

2. Find the limiting reagent to how much hydrogen is formed:

Take the number of moles of each compound and divide it between the stoichiometric coefficient in the balanced chemical equation.

- For the Mg:

`(0.100)/(1)`

- For the HCl:

`(0.500)/(2)=0.25`

The magnesium is the limiting reagent.

3. Find the number of moles of hydrogen produced:

`0.100molesMg(1molH_(2))/(1molMg)=0.100molesH_(2)`

4. Find the volume of hydrogen using the ideal gas equation at STP:

The ideal gas equation is expressed as
`PV=nRT`, where P is the pressure, at STP P=1 atm, V is the volume, n is the number of moles, R is a constante whose value is R=0.08206
`(L.atm)/(K.mol)`, and T is the temperature, at STP T=273K

Solving for V:

`V=(nRT)/(P)`

Replacing values:

`V=(0.100moles*0.08206(L.atm)/(K.mol)*273K)/(1atm)`

`V=2.24L` of
`H_(2)`