A 0.250-g sample of a magnesium-aluminum alloy dissolves completely in an excess of HCl (aq) . When the liberated H2 (g) is collected over water at 29 degrees celsius and 752 torr, the volume is found to be 301 mL . The vapor pressure of water at 29 degrees celsius is 30.0 torr.
Part A
How many moles H2 of can be produced from x grams of Mg in magnesium-aluminum alloy? The molar mass of Mg is 24.31 g/mol .
Part B
How many moles of H2 can be produced from y grams of Al in magnesium-aluminum alloy? The molar mass of Al is 26.98 g/mol.
How many moles of H2 can be produced from x grams of Mg in magnesium-aluminum alloy? The molar mass of Mg is 24.31 g/mol.
x g Mg×1 mol Mg/24.3050 g Mg×1 mol H2/1 mol Mg=0.0411x mol H2
0.0411x mol H2
How many moles of H2 can be produced from y grams of Al in magnesium-aluminum alloy? The molar mass of Al is 26.98 g/mol.
y g Al×1 mol Al/26.982 g Al×3 mol H2/2 mol Al=0.0556y mol H2
0.0556y mol H2
Really? This is your equation for part A? **Please do not use this or it will be wrong..
Rachel is right
hm ok anyways
Works for me
thank u rachel
To answer Part A, we need to determine the moles of H2 produced from the given amount of magnesium (Mg). We can calculate this using the stoichiometry of the reaction between Mg and HCl.
Here's the balanced equation for the reaction:
Mg (s) + 2HCl (aq) -> MgCl2 (aq) + H2 (g)
Step 1: Calculate moles of Mg
Given: Mass of Mg = x grams, Molar mass of Mg = 24.31 g/mol
Moles of Mg = (Mass of Mg) / (Molar mass of Mg)
Moles of Mg = x grams / 24.31 g/mol
Step 2: Use stoichiometry to determine moles of H2
From the balanced equation, we can see that 1 mole of Mg reacts with 1 mole of H2.
So the moles of H2 produced will be equal to the moles of Mg: Moles of H2 = Moles of Mg
Therefore, the moles of H2 that can be produced from x grams of Mg is (x grams / 24.31 g/mol).
Now let's move on to Part B to determine the moles of H2 produced from y grams of Al.
Step 3: Calculate moles of Al
Given: Mass of Al = y grams, Molar mass of Al = 26.98 g/mol
Moles of Al = (Mass of Al) / (Molar mass of Al)
Moles of Al = y grams / 26.98 g/mol
Step 4: Use stoichiometry to determine moles of H2
From the balanced equation, we can see that 1 mole of Al reacts with 3 moles of H2.
So the moles of H2 produced will be three times the moles of Al: Moles of H2 = 3 * Moles of Al
Therefore, the moles of H2 that can be produced from y grams of Al is 3 * (y grams / 26.98 g/mol).
Remember to substitute the values of x and y into the respective formulas to find the final answers.
I think there is a lot of extraneous material in the problem. That may be confusing you.
A. Mg + 2HCl ==> MgCl2 + H2
moles Mg = x/24.32.
Using the coefficients in the balanced equation, convert moles Mg to moles H2
(x/24.32) x (1 mole H2/1 mol Mg)
Now convert moles H2 to grams H.
(x/24.32) x (1 molH2/1 molMg) x (2 g H/1 mol H2) = ??
B is done the same way but the numbers will be different since it is a different equation.
I'll let you finish.