excess magnesium is reacted with 45 grams of sulfuric acid. 9.5 liters of hydrogen gas are collected understand conditions of temperature and pressure. what is the percentage of hydrogen sulfate and of water in the acod solution?
Mg + H2SO4 ==> MgSO4 + H2
mols H2 = 9.5L/22.4L = 0.5 ESTIMATED so you should redo the calculation as well as all that follow for they are estimates also.
0.5 mol H2 means you started with 0.5 mol H2SO4
g H2SO4 = mols H2SO4 x molar mass H2SO4 = estimated 42 grams
The solution had a mass of 45g, 42 of that was H2SO4 so 3 must have been H2O.
%H2SO4 = (42/45)*100 = ?
%H2O = (3/45)*100 = ?
Remember to recalculate all of these numbers.
To determine the percentage of hydrogen sulfate and water in the acid solution, we need to calculate the amount of hydrogen sulfate formed in the reaction.
Let's start by writing the balanced chemical equation for the reaction between excess magnesium (Mg) and sulfuric acid (H₂SO₄):
Mg + H₂SO₄ -> MgSO₄ + H₂↑
From the balanced equation, we can see that one mole of magnesium (Mg) reacts with one mole of sulfuric acid (H₂SO₄) to produce one mole of magnesium sulfate (MgSO₄) and one mole of hydrogen gas (H₂).
1 mole of H₂SO₄ = 2 moles of H (from the H₂)
So, the molar ratio of H₂SO₄ to H₂ is 1:2.
First, let's calculate the moles of hydrogen gas produced.
We can use the ideal gas equation, PV = nRT, where:
P is the pressure (unknown in this case),
V is the volume of gas (9.5 liters),
n is the number of moles of gas (H₂),
R is the ideal gas constant (0.0821 L·atm/(mol·K)),
T is the temperature (unknown in this case).
Rearranging the equation to solve for n (number of moles):
n = PV / RT
Since we don't have the pressure and temperature, we need more information to calculate the number of moles of hydrogen gas accurately.
To determine the percentage of hydrogen sulfate and water in the acid solution, we need to know the concentrations or volumes of the acid and water in the final solution after the reaction.