calculate the volume of hydrogen produced when 6g of magnesium is reacted with 100cm of 1 mol l-1 hydrochloric acid
no of moles of Mg= mass/gfm = 6/24.3 = 0.247 moles
1/2 mole of Mg= 1/2 mole of H2
= 0.247 moles x 100 = 24.7cm3 ???
This is a limiting reagent (LR) problem and we know that because amounts are given for BOTH reactants. I do these the long way.
mols Mg = your number is right at 0.247 and that will produce 0.247 mols H2 if you had all of the HCl needed.
mols HCl = M x L = 1 x 0.1 = 0.1
0.1 mols HCl x (1 mol H2/2 mols HCl) = 0.1 x 1/2 = 0.0.05 mols H2.
These two values for mols H2 don't agree; the answer in LR problems is ALWAYS the smaller value; therefore, you will produce 0.05 mols H2.
1 mol occupies 22.4 L at STP.
0.05 mols x 22.4 L/mol = ?
You don't list any conditions; you must mean at STP.
Thank you that was really helpful Dr Bob
To calculate the volume of hydrogen gas produced, we need to use the molar ratio between magnesium and hydrogen gas produced during the reaction.
The balanced chemical equation for the reaction between magnesium and hydrochloric acid is as follows:
Mg + 2HCl -> MgCl2 + H2
From the equation, we can see that for every 1 mole of magnesium (Mg) consumed, 1 mole of hydrogen gas (H2) is produced.
Given that the number of moles of magnesium (Mg) is 0.247 moles, we can determine the number of moles of hydrogen gas produced using the molar ratio:
moles of H2 = moles of Mg
So, the number of moles of hydrogen gas produced is also 0.247 moles.
Now, we can use the ideal gas equation, PV = nRT, to calculate the volume of hydrogen gas produced.
P: Pressure (atm) - Since the pressure is not given, we will assume it to be at standard atmospheric pressure, which is approximately 1 atm.
V: Volume (L) - This is what we need to calculate.
n: Number of moles (mol) - 0.247 moles.
R: Ideal gas constant - 0.0821 L·atm/(mol·K).
T: Temperature (K) - again, not provided, so we will assume it to be room temperature, 298 K.
Using the ideal gas equation and rearranging for the volume (V), we have:
V = (n * R * T) / P
Plugging in the values we have:
V = (0.247 mol * 0.0821 L·atm/(mol·K) * 298 K) / 1 atm
Calculating:
V = 6.06 L
Therefore, the volume of hydrogen gas produced when 6g of magnesium is reacted with 100 cm3 of 1 mol/L hydrochloric acid is 6.06 L.
To calculate the volume of hydrogen gas produced when 6g of magnesium is reacted with 100cm³ of 1 mol/L hydrochloric acid, you need to consider the balanced chemical equation for the reaction between magnesium and hydrochloric acid:
Mg + 2HCl -> MgCl₂ + H₂
According to the equation, 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas.
First, calculate the number of moles of magnesium using its molar mass (gfm - grams formula mass). The molar mass of magnesium is 24.3 g/mol.
Number of moles of magnesium = mass / gfm = 6g / 24.3 g/mol = 0.247 moles
Since 1 mole of magnesium produces 1 mole of hydrogen gas, the number of moles of hydrogen gas produced would also be 0.247 moles.
Now, we have to convert the moles of hydrogen gas to volume using the ideal gas law, which states that 1 mole of any ideal gas occupies 22.4 liters (or 22,400 cm³) at standard temperature and pressure (STP).
Volume of hydrogen gas = number of moles x molar volume at STP
Volume of hydrogen gas = 0.247 moles x 22,400 cm³/mol = 5,528.8 cm³
Therefore, the volume of hydrogen gas produced when 6g of magnesium is reacted with 100cm³ of 1 mol/L hydrochloric acid is approximately 5,528.8 cm³.