zinc metal react with HCI to give zinc chloride and hydrogen if 260g of zinc reacts with excess HCI, what's the volume of hydrogen produced at RIP?
200g of Zn is 3.06 moles
so you will get 3.06 moles of H2
That would be 22.4L/mole at STP
No idea what it is at RIP ...
thanks for the help and sorry spelling mistake its STP.
To calculate the volume of hydrogen produced at RTP (room temperature and pressure), you need to use the Ideal Gas Law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Let's go through the steps to find the volume of hydrogen gas produced:
1. Find the number of moles of zinc (Zn) using its molar mass.
- The molar mass of zinc (Zn) is approximately 65.38 g/mol.
- Divide the given mass of zinc (260 g) by its molar mass:
260 g / 65.38 g/mol = 3.976 mol
2. Determine the stoichiometry of the reaction.
- According to the balanced equation, the reaction between zinc and hydrochloric acid (HCl) is as follows:
Zn + 2HCl -> ZnCl2 + H2
- For every 1 mole of zinc (Zn) reacted, 1 mole of hydrogen gas (H2) is produced.
3. Calculate the number of moles of hydrogen gas produced.
- Since the molar ratio between zinc and hydrogen gas is 1:1, the number of moles of hydrogen gas is also 3.976 mol.
4. Use the Ideal Gas Law equation to find the volume of hydrogen gas at RTP.
- Rearrange the equation to solve for volume (V):
V = (nRT) / P
Here, you need to know the value of the ideal gas constant (R) and the temperature (T) at RTP.
- The ideal gas constant, R, is approximately 0.0821 L·atm/(mol·K).
- The temperature at RTP is typically taken as 273.15 K.
- Assuming the pressure (P) is 1 atm (since the problem states RTP), we can substitute the values:
V = (3.976 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
Calculating the value gives us:
V ≈ 89.625 L or approximately 89.6 L
Therefore, the volume of hydrogen gas produced at RTP (room temperature and pressure) is approximately 89.6 liters.