2K (s) + 2H2O (l) --> 2KOH (aq) + H2 (g)
When 0.400 mole of potassium reacts with excess water at STP, what is the volume of hydrogen gas produced?
the answer is 4.48 liters but I keep getting wrong numbers!
please help!
Thanks!
From the equation 2 moles of K react to form 1 mole of H2 gas. At STP 1 mole of any gas occupies 22.4 litres.
0.400 mol of K will give 0.200 mole of H2 gas.
Volume = 0.200 mole x 22.4 l mole^-1
Thanks!!
To find the volume of hydrogen gas produced, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure (in this case, at STP, it is 1 atm)
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L.atm/mol.K)
T is the temperature (in this case, at STP, it is 273 K)
First, let's calculate the number of moles of hydrogen gas produced. According to the balanced chemical equation, 2 moles of potassium (K) produce 1 mole of hydrogen gas (H2). Therefore, if 0.400 moles of potassium react, half of it, or 0.200 moles, will be converted into hydrogen gas.
Using the ideal gas law equation, we can rearrange it to solve for volume:
V = nRT / P
V = (0.200 mol) * (0.0821 L.atm/mol.K) * (273 K) / (1 atm)
V ≈ 4.48 L
So, the volume of hydrogen gas produced is approximately 4.48 liters.
To find the volume of hydrogen gas produced, you can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
In this case, you have the number of moles of potassium (0.400 mol) and the reaction is carried out at standard temperature and pressure (STP). At STP, the temperature is 273.15 K, and the pressure is 1 atm.
First, you need to calculate the number of moles of hydrogen gas produced in the reaction by using the stoichiometry of the balanced equation. From the balanced equation, you can see that 2 moles of potassium (2K) react to produce 1 mole of hydrogen gas (H2).
So, since 2 moles of potassium produce 1 mole of hydrogen gas, 0.400 moles of potassium will produce (0.400 mol * 1 mol H2) / (2 mol K) = 0.200 moles of hydrogen gas.
Now, plug in the values into the ideal gas law equation:
PV = nRT
Rearrange the equation to solve for V:
V = (nRT) / P
Substitute the values:
V = (0.200 mol * 0.0821 L·atm/mol·K * 273.15 K) / (1 atm)
V = 4.429 L
Rounding to the correct number of significant figures, the volume of hydrogen gas produced is 4.43 liters, not 4.48 liters as mentioned previously.
Please double-check your calculations to ensure you're using the correct values and following the steps correctly.