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.
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
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.
Well, well, well! Looks like we have a situation here involving some chemistry! I must say, this is definitely my element...pun intended! Let's dive into it, shall we?
The molar volume of a gas at S.T.P is indeed 22.4 dm^3. However, in this case, we are dealing with 576 cm^3 of hydrogen. No worries, I'll handle the unit conversion! *bam-bam!* Drumroll, please!
Alrighty, here we go: 1 dm^3 = 1000 cm^3. So, to convert 576 cm^3 to dm^3, we divide it by 1000.
576 cm^3 ÷ 1000 = 0.576 dm^3
Ta-da! We now have the volume in the right units, and it's 0.576 dm^3. Hooray for conversions! But we're not done yet, my friend. Oh no!
You mentioned that an excess of a divalent metal M was used. Since we know the molar volume of the gas, we can determine the number of moles of hydrogen gas that were produced. And once we have the number of moles, we can calculate the mass of the metal using its molar mass.
The equation for the reaction between the metal M and hydrochloric acid is:
M + 2HCl → MCl2 + H2
From the equation, we can see that 1 mole of metal M produces 1 mole of hydrogen gas.
Since the volume of hydrogen gas produced is 0.576 dm^3, we can calculate the number of moles using the molar volume:
Number of moles = Volume ÷ Molar volume
Number of moles = 0.576 dm^3 ÷ 22.4 dm^3/mol
And now, with a little math magic...
Number of moles = 0.0257 moles
Voila! We have the number of moles of hydrogen gas produced. But what about the mass of the metal M? Well, hold your nose, because we're about to take a dive into the periodic table!
The molar mass of M is given as 24 g/mol. So, to find the mass of M, we multiply the number of moles by the molar mass:
Mass of M = Number of moles × Molar mass
Mass of M = 0.0257 moles × 24 g/mol
Drumroll, please... *ba-dum-tss*
Mass of M = 0.6168 g
And there you have it! The mass of the metal M that produced this volume of hydrogen gas is approximately 0.6168 grams. So, keep those chemistry hats on, my friend, and keep exploring the wonderful world of reactions and conversions!
To find the mass of the metal that produced the volume of hydrogen, we need to use the stoichiometry of the reaction between the metal M and hydrochloric acid.
The balanced chemical equation for the reaction between metal M and hydrochloric acid is:
M + 2HCl → MCl₂ + H₂
From the equation, we can see that 1 mole of metal M reacts with 2 moles of HCl and produces 1 mole of hydrogen gas.
Given that 576 cm^3 of hydrogen gas is produced at STP (standard temperature and pressure), we need to convert this volume into moles of hydrogen gas.
1 mole of any gas at STP occupies a volume of 22.4 dm^3, so we can use this conversion factor to convert cm^3 to moles.
1 cm^3 = 1/1000 dm^3
Hence, the volume of hydrogen gas in dm^3 can be calculated as follows:
576 cm^3 × (1/1000) dm^3/cm^3 = 0.576 dm^3
Now we can use the molar ratio from the balanced equation to calculate the number of moles of hydrogen gas.
1 mole of hydrogen gas = 0.576 dm^3 of hydrogen gas
Since the molar ratio between the metal M and hydrogen gas is 1:1, this means that 0.576 moles of metal M were used in the reaction.
To calculate the mass of metal M, we need to use its molar mass, which is given as 24 g/mol.
Mass of metal M = moles of metal M × molar mass of metal M
= 0.576 moles × 24 g/mol
= 13.824 g
Therefore, the mass of the metal that produced 576 cm^3 of hydrogen gas is 13.824 grams.