High School

We appreciate your visit to In the reaction between 1 74 g of zinc and excess hydrogen chloride how many liters of gas hydrogen will be formed at 305 K. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

In the reaction between 1.74 g of zinc and excess hydrogen chloride, how many liters of gas (hydrogen) will be formed at 305 K and 98.3 kPa?

Answer :

To solve this problem, we need to use the balanced chemical equation for the reaction between zinc and hydrogen chloride, and then apply the appropriate gas law calculations to determine the volume of hydrogen gas formed.

Given information:
- Mass of zinc (Zn) = 1.74 g
- Temperature (T) = 305 K
- Pressure (P) = 98.3 kPa

Step 1: Write the balanced chemical equation for the reaction.
Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

Step 2: Calculate the number of moles of zinc (Zn) using the given mass.
Moles of Zn = Mass of Zn / Molar mass of Zn
Moles of Zn = 1.74 g / (65.38 g/mol)
Moles of Zn = 0.0266 mol

Step 3: Use the balanced equation to determine the number of moles of hydrogen (H2) produced.
Moles of H2 = Moles of Zn × 1 (from the balanced equation)
Moles of H2 = 0.0266 mol

Step 4: Calculate the volume of hydrogen gas using the ideal gas law:
PV = nRT
V = (nRT) / P
V = (0.0266 mol × 8.314 J/(mol·K) × 305 K) / (98.3 kPa)
V = 2.41 L

Therefore, 2.41 liters of hydrogen gas will be formed at 305 K and 98.3 kPa.

Thanks for taking the time to read In the reaction between 1 74 g of zinc and excess hydrogen chloride how many liters of gas hydrogen will be formed at 305 K. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada