Question

When magnesium reacts with hydrochloric acid, hydrogen gas is formed: 2HCl + Mg → H2 + MgCl2. What is the volume of hydrogen produced at 25°C and 101.3 kilopascals when 49.0 grams of HCl reacts with excess magnesium? Use the periodic table and ideal gas resource. A. 1.38 L B. 2.76 L C. 16.4 L D. 32.9 L E. 33.1 L

Answer 1

Answer:- C. 16.4 L

Solution:- The given balanced equation is:

$2HCl+Mg\rightarrow H_2+MgCl_2$

From this equation, there is 2:1 mol ratio between HCl and hydrogen gas. First of all we calculate the moles of hydrogen gas from given grams of HCl using stoichiometry and then the volume of hydrogen gas could be calculated using ideal gas law equation, PV = nRT.

Molar mass of HCl = 1.008 + 35.45 = 36.458 gram per mol

The calculations are shown below:

$49.0gHCl(\frac{1molHCl}{36.458gHCl})(\frac{1molH_2}{2molHCl})$

= $0.672molH_2$

Now we will use ideal gas equation to calculate the volume.

n = 0.672 mol

T = 25 + 273 = 298 K

P = 101.3 kPa = 1 atm

R = $0.0821\frac{atm.L}{mol.K}$

PV = nRT

1(V) = (0.672)(0.0821)(298)

V = 16.4 L

From calculations, 16.4 L of hydrogen gas are formed and so the correct choice is C.