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tester [92]
3 years ago
11

How many liters of h2 gas, collected over water at an atmospheric pressure of 752 mmhg and a temperature of 21.0°c, can be made

from 1.566 g of zn and excess hcl? the partial pressure of water vapor is 18.65 mmhg at 21.0°c?
Chemistry
2 answers:
sleet_krkn [62]3 years ago
5 0

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

Shalnov [3]3 years ago
3 0

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

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