What volume of hydrogen would be produced by the reaction of 2.00g of Mg with an excess of hydrochloric acid at STP. Using this equation Mg+HCl
Mg + 2HCl ==> MgCl2 + H2
Convert 2.00 g Mg to moles. moles = grams/molar mass.
Using the coefficients in the balanced equation, convert moles Mg to moles H2.
Now convert moles H2 to volume. moles x 22.4 L/mol = ??L
1.84
To determine the volume of hydrogen gas produced, we need to use the balanced chemical equation and the ideal gas law.
The balanced chemical equation for the reaction between magnesium (Mg) and hydrochloric acid (HCl) is:
Mg + 2HCl → MgCl2 + H2
From the equation, we can see that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas.
Step 1: Calculate the number of moles of magnesium (Mg):
Given the mass of magnesium (2.00 g) and the molar mass of magnesium (24.31 g/mol), we can calculate the number of moles.
Number of moles (n) = mass / molar mass
n = 2.00 g / 24.31 g/mol
n ≈ 0.0822 mol
Step 2: Calculate the volume of hydrogen gas produced using the ideal gas law:
The ideal gas law equation is: PV = nRT
where:
P = pressure (STP conditions: 1 atm)
V = volume
n = number of moles
R = ideal gas constant (0.0821 L * atm / mol * K)
T = temperature (STP conditions: 273 K)
Rearranging the ideal gas law equation to solve for the volume (V):
V = nRT / P
V = (0.0822 mol)(0.0821 L * atm / mol * K)(273 K / 1 mol) / 1 atm
V ≈ 2.20 L
Therefore, approximately 2.20 liters of hydrogen gas would be produced by the reaction of 2.00 g of Mg with an excess of hydrochloric acid at STP.
To determine the volume of hydrogen gas produced by the reaction, you need to use the concept of stoichiometry. Stoichiometry allows you to relate the amounts of reactants and products in a chemical equation.
1. Begin by writing and balancing the chemical equation:
Mg + 2HCl → MgCl2 + H2
2. Convert the given mass of Mg (2.00g) to moles:
The molar mass of Mg is 24.31 g/mol.
Number of moles of Mg = 2.00g / 24.31 g/mol
3. Use the balanced chemical equation and stoichiometry to determine the mole ratio between Mg and H2. From the equation, you can see that 1 mole of Mg reacts to form 1 mole of H2.
4. The volume of 1 mole of any gas at STP (Standard Temperature and Pressure) is 22.4 liters. Therefore, if you determine the number of moles of H2 produced, you can directly convert it to the volume.
Now, let's calculate the volume of hydrogen gas produced:
1. Determine the number of moles of Mg:
Number of moles of Mg = 2.00g / 24.31 g/mol
2. Using the mole ratio from the balanced equation:
Number of moles of H2 = Number of moles of Mg
3. Convert moles of H2 to volume at STP:
Volume of H2 gas = Number of moles of H2 * 22.4 L/mol
By following these steps, you can calculate the volume of hydrogen gas produced.