Answer: 0.299 liter
Explanation:
1) Balanced chemical equation
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g) ↑
2) Mole ratios
1 mol Zn(s) : 2 mol HCl(aq) : 1mol ZnCl₂(aq) : 2 mol H₂(g)
3) Convert 1.566 g Zn into number of moles (n)
n = 1.566g / 65.39 g/mol = 0.02395 mol
4) Set the proportion
1 mol Zn / 2 mol H₂ = 0.02395 mol Zn / x ⇒ x = 0.01197 mol H₂
5) Calcualte the partial pressure of H₂
Partial pressure H₂ = Total pressure - partial pressure of water vapor
Partial pressure H₂ = 752 mmHg - 18.65 mmHg = 733.35 mmHg
Convert to atm = 733.5 mmHg × 1 atm/ 760mmHg = 0.9649 atm
6) Calculate volume using ideal gas equation
PV = nRT ⇒ V = nRT/P
⇒ V = 0.01197 mol × 0.0821 atm×liter / k×mol × (21 + 273k) / 0.9649atm
V = 0.2994 liter
The answer is V = 0.6 L
The explanation:
1- First we will get P (H) :
p (H) = p total - p (H2O)
= 752 mmHg - 18.65 mmHg
= 733 mmHg
= 733 mmHg X (1 atm / 760 atm) = 0.964 atm
2- then when the reaction equation is:
Zn(s)+2HCl(aq) -> H2(g)+ZnCl2(aq)
so, then 1 mole of Zn will produce 1 mole of H2.
so, we need to get first the mole of Zn:
mole Zn = mass / molar mass
= 1.566g / 65.38g/mol
= 0.02395mol
∴ 0.02395 mol of Zn will produce 0.02395 mol of H2
by using the ideal gas equation we can get the volume V :
when :
PV = nRT
when T is the temperature = 21 + 273 = 294 K
and R is the ideal gas constant = 0.0821 L atm /K mol
and n is the number of moles = 0.2395 mol
and P is the pressure = 0.964 atm
so, by substitution:
V = (0.02395 moles H2)(0.0821 L atm / K mole)(294 K) / (0.964 atm)
= 0.6 L
