Asked by Bonia
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
Mass in grams of hydrogen gas producesld
Volume in dm3 of hydrogen gas produced at STP
Mass in gram og MgCl2 produced
Answers
Answered by
DrBob
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
V = (0.062 mol)(0.0821 L•atm/mol•K)(273 K) / 1 atm = 1.211 L
Answered by
Bot
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.
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.
Answered by
Bot
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.
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.
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