How many liters of H2 gas can be produced at 0 ∘C and 1.00 atm (STP) from 10.0 g of Zn?
Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)
each mole of zinc produces a mole of hydrogen gas
a mole of gas occupies 22.4 L at STP
To find the number of liters of H2 gas produced, we can use the ideal gas law.
The balanced equation tells us that 1 mole of Zn will produce 1 mole of H2 gas.
First, we need to convert the given mass of Zn to moles.
1. Calculate the molar mass of Zn:
Zn = 65.38 g/mol
2. Use the following equation to calculate the number of moles of Zn:
moles of Zn = mass of Zn / molar mass of Zn
moles of Zn = 10.0 g / 65.38 g/mol
Next, we can use the stoichiometry from the balanced equation to determine the number of moles of H2 gas produced.
From the balanced equation, we see that 1 mole of Zn produces 1 mole of H2 gas.
Therefore, the number of moles of H2 gas produced is equal to the number of moles of Zn.
Now, we can use the ideal gas law to calculate the volume of gas produced.
The ideal gas law equation is:
PV = nRT
Where:
P = pressure (1.00 atm)
V = volume (unknown)
n = number of moles (from above)
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature (0 °C = 273 K)
Rearranging the equation to solve for V:
V = (nRT) / P
Plug in the values:
V = (moles of H2 gas) x R x T / P
V = (moles of Zn) x R x T / P
3. Convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273
T = 0°C + 273 = 273 K
Plug in the values:
V = (10.0 g / 65.38 g/mol) x (0.0821 L·atm/(mol·K)) x (273 K) / 1.00 atm
Now, we can calculate the volume of H2 gas produced:
V = (10.0 g / 65.38 g/mol) x (0.0821 L·atm/(mol·K)) x (273 K) / 1.00 atm
V ≈ 4.03 L
Therefore, approximately 4.03 liters of H2 gas can be produced.
To determine the number of liters of H2 gas that can be produced from 10.0 g of Zn, you need to use stoichiometry and the ideal gas law.
1. Start by balancing the chemical equation:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
2. Determine the moles of zinc (Zn) from its mass:
To find the moles of Zn, divide the given mass by its molar mass. The molar mass of Zn is 65.38 g/mol.
Moles of Zn = 10.0 g / 65.38 g/mol
3. Use the stoichiometric ratio between Zn and H2 to find the moles of H2 produced:
From the balanced chemical equation, we see that 1 mole of Zn produces 1 mole of H2.
So, the moles of H2 produced will be the same as the moles of Zn.
4. Apply the ideal gas law to calculate the volume of H2:
The ideal gas law is given by 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.
At STP (standard temperature and pressure), the values are:
P = 1.00 atm
T = 0 °C = 273.15 K
R = 0.0821 L·atm/(mol·K) (the ideal gas constant)
Plug in these values, along with the number of moles of H2, into the ideal gas law:
V = (nRT) / P
Calculate the volume of H2 gas in liters.
5. Finally, convert the volume of H2 gas from liters to the desired units (if necessary).
Note: Make sure to use consistent units throughout the calculations to get the accurate result.
By following these steps, you can determine how many liters of H2 gas can be produced from 10.0 g of Zn.