Question

How many grams of water are required to produce 5.50 L of hydrogen gas at 25.0°C and 755 mm Hg pressure according to the chemical equation shown below?

Answer 1

Answer is: 4.02 grams of water are required.
Chemical reaction: BaH₂ + 2H₂O → Ba(OH)₂ + 2H₂.
Ideal gas law: p·V = n·R·T.
p = 755 mm Hg ÷ 760.0 mmHg / atm = 0.993 atm.
T = 25 + 273.15 = 298.15 K.
V(H₂) = 5.50 L.
R = 0,08206 L·atm/mol·K.
n(H₂) = 0.993 atm · 5.5 L ÷ 0,08206 L·atm/mol·K · 298.15 K.
n(H₂) = 0.223 mol.
From chemical reaction: n(H₂O) : n(H₂) = 1 : 1.
n(H₂O) = 0.223 mol.
m(H₂O) = 0.223 mol · 18 g/mol.
m(H₂O) = 4.02 g.

Answer 2

Answer:

4.014 g of water

Explanation:

An internet search for your question tells me that the chemical equation is:

• BaH₂(s) + 2H₂O(l)  → Ba(OH)₂(aq) + 2H₂(g)

So first we use PV=nRT to calculate the moles of hydrogen gas produced:

• P = 755 mmHg ⇒ 755/760 = 0.993 atm

• T = 25 °C ⇒ 25 + 273.15 = 298.16 K

• V = 5.50 L

0.993 atm * 5.50 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K

• n = 0.223 mol H₂

Now we convert mol H₂ to mol H₂O and finally to grams of water:

0.223 mol H₂ * $\frac{1molH_{2}O}{1molH_{2}}*\frac{18g}{1molH_{2}O}$ = 4.014 g H₂O