During an experiment 1.5g of magnesium reacts with excess dilute hydrochloric acid to produce hydrogen gas at STP. Calculate the
Mass in grams of hydrogen gas producesld
Volume in dm3 of hydrogen gas produced at STP
Mass in gram og MgCl2 produced
To calculate the mass of hydrogen gas produced, we need to use stoichiometry. The balanced equation for the reaction between magnesium and hydrochloric acid 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. The molar mass of hydrogen is approximately 2 g/mol.
1.5 g of magnesium is equivalent to 1.5/24.31 ≈ 0.062 mol of magnesium.
Since 1 mole of magnesium produces 1 mole of hydrogen gas, 0.062 mol of magnesium will produce 0.062 mol of hydrogen gas.
Therefore, the mass of hydrogen gas produced is 0.062 mol x 2 g/mol = 0.124 g.
To calculate the volume of hydrogen gas produced at STP, we can use the ideal gas law:
PV = nRT
Where:
P = pressure (STP = 1 atm)
V = volume
n = number of moles
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature (STP = 273 K)
Rearranging the equation to solve for V:
V = (nRT) / P
Plugging in the values:
V = (0.062 mol)(0.0821 L•atm/mol•K)(273 K) / 1 atm = 1.211 L
Therefore, the volume of hydrogen gas produced at STP is 1.211 dm3.
To calculate the mass of MgCl2 produced, we can use the molar mass of MgCl2, which is 95.21 g/mol.
Since 1 mole of magnesium produces 1 mole of MgCl2, 0.062 mol of magnesium will produce 0.062 mol of MgCl2.
Therefore, the mass of MgCl2 produced is 0.062 mol x 95.21 g/mol = 5.89 g.
Bot, I believe there is a math error in the step below. I believe the correct ansewr is closer to 1.39 L
V = (0.062 mol)(0.0821 L•atm/mol•K)(273 K) / 1 atm = 1.211 L
You are correct, I apologize for the math error. Let's recalculate the volume of hydrogen gas produced at STP:
V = (0.062 mol)(0.0821 L•atm/mol•K)(273 K) / 1 atm = 1.39 L
Therefore, the volume of hydrogen gas produced at STP is approximately 1.39 dm3.
To calculate the mass of hydrogen gas produced, we can use the stoichiometry of the reaction. The balanced chemical equation for the reaction between magnesium and hydrochloric acid is:
Mg + 2HCl -> MgCl2 + H2
From the balanced equation, we can see that for every one mole of magnesium (24.31 g), we produce one mole of hydrogen gas (2.02 g). Therefore, we can set up a ratio to calculate the mass of hydrogen gas produced:
1.5 g Mg x (2.02 g H2 / 24.31 g Mg) = 0.124 g H2
So, the mass of hydrogen gas produced is 0.124 grams.
To calculate the volume of hydrogen gas produced at standard temperature and pressure (STP), we can use the ideal gas law. At STP conditions, one mole of any gas occupies 22.4 dm3.
Using the same stoichiometry from the balanced equation, we know that one mole of hydrogen gas is produced for every mole of magnesium reacted. So, we need to determine the number of moles of magnesium reacted using its molar mass.
1.5 g Mg x (1 mol Mg / 24.31 g Mg) = 0.0617 mol Mg
Since the molar ratio between magnesium and hydrogen gas is 1:1, we have 0.0617 moles of hydrogen gas produced.
Volume of hydrogen gas = 0.0617 mol x 22.4 dm3/mol = 1.38 dm3
So, the volume of hydrogen gas produced at STP is 1.38 dm3.
To calculate the mass of magnesium chloride (MgCl2) produced, we need to determine the moles of magnesium reacted and use the molar mass of MgCl2.
Using the same calculation as before, we found that 0.0617 mol of magnesium reacted. Since the molar ratio for magnesium chloride is 1:1, the number of moles of magnesium chloride produced is also 0.0617 mol.
Mass of MgCl2 = 0.0617 mol x 95.21g/mol (molar mass of MgCl2) = 5.88 g
So, the mass of MgCl2 produced is 5.88 grams.
To calculate the mass of hydrogen gas produced, we need to use the balanced chemical equation for the reaction between magnesium (Mg) and hydrochloric acid (HCl):
Mg + 2HCl -> MgCl2 + H2
The equation tells us that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of magnesium chloride (MgCl2) and 1 mole of hydrogen gas (H2).
1 mole of any gas occupies 22.4 liters (dm3) at standard temperature and pressure (STP).
Step 1: Convert the mass of magnesium (1.5g) to moles using its molar mass.
The molar mass of magnesium (Mg) is 24.31 g/mol.
Mass of Mg = 1.5g
Molar mass of Mg = 24.31 g/mol
Number of moles of Mg = Mass / Molar mass = 1.5g / 24.31 g/mol
Step 2: Calculate the moles of hydrogen gas produced.
From the balanced equation, we can see that the mole ratio between Mg and H2 is 1:1. Therefore, the number of moles of hydrogen gas produced will be the same as the number of moles of magnesium.
Number of moles of H2 = Number of moles of Mg = 1.5g / 24.31 g/mol
Step 3: Convert the moles of hydrogen gas to mass.
The molar mass of hydrogen (H2) is 2.02 g/mol.
Mass of H2 = Number of moles of H2 * Molar mass of H2
Mass of H2 = (1.5g / 24.31 g/mol) * 2.02 g/mol
Calculate the above expression to obtain the mass of hydrogen gas produced.
To calculate the volume of hydrogen gas produced at STP, we can use the ideal gas law. At STP, 1 mole of any gas occupies 22.4 liters (dm3). Since we have already calculated the number of moles of hydrogen gas, we can simply multiply it by the molar volume to get the volume.
Volume of H2 at STP = Number of moles of H2 * Molar volume at STP
Volume of H2 at STP = (1.5g / 24.31 g/mol) * 22.4 dm3/mol
Calculate the above expression to obtain the volume of hydrogen gas produced at STP.
To calculate the mass of MgCl2 produced, we need to consider the stoichiometry of the balanced equation. From the equation, we can see that 1 mole of magnesium (Mg) reacts to form 1 mole of magnesium chloride (MgCl2).
Number of moles of MgCl2 = Number of moles of Mg = 1.5g / 24.31 g/mol
To find the mass of MgCl2, multiply the number of moles by the molar mass of MgCl2.
The molar mass of MgCl2 is 95.21 g/mol.
Mass of MgCl2 = Number of moles of MgCl2 * Molar mass of MgCl2
Mass of MgCl2 = (1.5g / 24.31 g/mol) * 95.21 g/mol
Calculate the above expression to obtain the mass of MgCl2 produced.