Respuesta :
Answer:
Number of moles of sodium reacted = 0.707 moles
Explanation:
P(H₂) = P(T) – P(H₂O)
P(H₂) = 754 – 17.5 = 736.5 mm Hg
Use the ideal gas equation which
PV= nRT, where P is the pressure V is the volume, n is the number of moles R is the Gas Constant and T is temperature
Re- arrange to calculate the number of moles and using the data provided
n = P x V/R x T
n =736.5 x 8.77/62.36367 x (mmHg/mol K) x (20 + 273)
n = 0.35348668
n = 0.353 moles H₂
from the equation we know that
0.353 mole H₂ x 2mole Na/1mole H₂, So
0.353 x 2 = 0.707 mole Na
The number of moles of Sodium metal reacted were 0.707 moles.
0.708 moles of Na react with water to produce 8.77 L of H₂ collected over water at 20 °C and 754 mmHg.
Sodium metal reacts with water to produce hydrogen gas according to the following equation:
2 Na(s) + 2 H₂O(l) ⇒ 2 NaOH(aq) + H₂(g)
Hydrogen is collected over water. The total pressure of the gaseous mixture is 754 mmHg. If the vapor pressure of water is 17.5 mmHg, the partial pressure of hydrogen is:
[tex]P = pH_2 + pH_2O\\\\pH_2 = P - pH_2O = 754 mmHg - 17.5 mmHg = 737 mmHg[/tex]
Then, we will convert 20 °C to Kelvin using the following expression.
[tex]K = \° C + 273.15 = 20\° C + 273.15 = 293 K[/tex]
Hydrogen occupies 8.77 L at 293 K and 737 mmHg. We can calculate the moles of hydrogen using the ideal gas equation.
[tex]P \times V = n \times R \times T\\\\n = \frac{P \times V}{ R \times T} = \frac{737mmHg \times 8.77L}{ (62.4mmHg/mol.K) \times 293K} = 0.354 mol[/tex]
The molar ratio of Na to H₂ is 2:1. The moles of Na that produced 0.354 moles of H₂ are:
[tex]0.354 mol H_2 \times \frac{2molNa}{1molH_2} = 0.708 mol Na[/tex]
0.708 moles of Na react with water to produce 8.77 L of H₂ collected over water at 20 °C and 754 mmHg.
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