An excess of a divalent metal M was dissolved in a limited volume of hydrochloric acid. If 576cm^3 of hydrogen were liberated at S.t.p , what was the mass of metal that produced this volume of hydrogen.(M=24,H=1, molar volume of a gas at s.t.p =22.4dm^3.
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M + 2HCl ≈ MCl2 + H2
This means that 1 mole of the metal produced 1 mole of H2
And 576cm³ of hydrogen was liberated at s.t.p
At s.t.p 1 mole of a gas is 22,400cm³ (22.4 dm³ = 22,400cm³)
Then at s.t.p y moles is 576 moles
Cross multiply
y × 22,400 = 1 × 576
Dividing through by 22,400
y = 0.02571 moles of hydrogen
And since 1 mole of metal liberated 1 mole of hydrogen
Therefore 0.02571 moles of the metal liberated 0.02571
Mass of the metal = no of moles × molar mass
= 0.02571 × 24
= 0.617g
This means that 1 mole of the metal produced 1 mole of H2
And 576cm³ of hydrogen was liberated at s.t.p
At s.t.p 1 mole of a gas is 22,400cm³ (22.4 dm³ = 22,400cm³)
Then at s.t.p y moles is 576 moles
Cross multiply
y × 22,400 = 1 × 576
Dividing through by 22,400
y = 0.02571 moles of hydrogen
And since 1 mole of metal liberated 1 mole of hydrogen
Therefore 0.02571 moles of the metal liberated 0.02571
Mass of the metal = no of moles × molar mass
= 0.02571 × 24
= 0.617g
15.0( g) of potassium trioxochlorate v was crushed and heated with about 0.1(g ) of magnesium IV oxide . write an equation of the reaction. Calculate the mass of c that would be produced assuming the reaction was complete . Fine the volume of oxygen produced by 1mole of potassium trioxochlorate(v )STP (o=16,cl=35.5,k=39)
The balanced equation for the reaction is:
2KClO3 + 3MgO → 3MgCl2 + 2KCl + 3O2
The reaction stoichiometry shows that for every 2 moles of KClO3, 3 moles of O2 are produced.
First, we need to calculate the moles of KClO3 used:
Moles of KClO3 = mass / molar mass = 15.0 g / (39.0 + (3 x 16.0) + 35.5) g/mol = 0.0917 mol
Next, we need to determine the limiting reagent. To do this, we compare the moles of KClO3 with the moles of MgO used:
Moles of MgO = mass / molar mass = 0.1 g / (24.3 + 16.0) g/mol = 0.0036 mol
The MgO is the limiting reagent because it is present in the smallest amount.
Now, we can use the stoichiometry of the balanced equation to calculate the mass of O2 produced:
From the balanced equation, 2 moles of KClO3 produce 3 moles of O2.
Therefore, 0.0917 moles of KClO3 will produce (3/2) x 0.0917 = 0.1376 moles of O2.
Mass of O2 = moles x molar mass = 0.1376 mol x 16.0 g/mol = 2.2016 g
Finally, we can use the molar volume of a gas at STP (22.4 L/mol) to calculate the volume of O2 produced by 1 mole of KClO3:
Volume of O2 = moles x molar volume = 0.1376 mol x 22.4 L/mol = 3.08 L/mol
Therefore, the volume of O2 produced by 1 mole of KClO3 at STP is 3.08 L.
2KClO3 + 3MgO → 3MgCl2 + 2KCl + 3O2
The reaction stoichiometry shows that for every 2 moles of KClO3, 3 moles of O2 are produced.
First, we need to calculate the moles of KClO3 used:
Moles of KClO3 = mass / molar mass = 15.0 g / (39.0 + (3 x 16.0) + 35.5) g/mol = 0.0917 mol
Next, we need to determine the limiting reagent. To do this, we compare the moles of KClO3 with the moles of MgO used:
Moles of MgO = mass / molar mass = 0.1 g / (24.3 + 16.0) g/mol = 0.0036 mol
The MgO is the limiting reagent because it is present in the smallest amount.
Now, we can use the stoichiometry of the balanced equation to calculate the mass of O2 produced:
From the balanced equation, 2 moles of KClO3 produce 3 moles of O2.
Therefore, 0.0917 moles of KClO3 will produce (3/2) x 0.0917 = 0.1376 moles of O2.
Mass of O2 = moles x molar mass = 0.1376 mol x 16.0 g/mol = 2.2016 g
Finally, we can use the molar volume of a gas at STP (22.4 L/mol) to calculate the volume of O2 produced by 1 mole of KClO3:
Volume of O2 = moles x molar volume = 0.1376 mol x 22.4 L/mol = 3.08 L/mol
Therefore, the volume of O2 produced by 1 mole of KClO3 at STP is 3.08 L.
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