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3 years ago

Stoichiometry (8A.S, 8B.R, 8G.S, 8H.S]:Question 4

Chemistry

1 answer:

Lostsunrise [7]3 years ago

4 0

Answer:

59.76 grams of Li₂O

Explanation:

To solve this problem we will first calculate the moles of Li₂O. One mole of given substance always contains 6.022 × 10²³ particles which can be atoms, ions, molecules or formula units. This number is also called as Avogadro's Number.

The relation between Moles, Number of Particles and Avogadro's Number is given as,

Number of Moles = Number of Particles ÷ 6.022 × 10²³

Putting values,

Number of Moles = 1.204 × 10²⁴ Particles ÷ 6.022 × 10²³

Number of Moles = 2.0 Moles

Secondly, we will convert calculated moles to mass using following relationship.

Moles = Mass / M.Mass

Or,

Mass = Moles × M.Mass

Putting values,

Mass = 2.0 mol × 29.88 g/mol

Mass = 59.76 grams of Li₂O

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Attempt 3 During an experiment, a student adds 2.90 g CaO 2.90 g CaO to 400.0 mL 400.0 mL of 1.500 M HCl 1.500 M HCl . The stude

kondor19780726 [428]

Answer:

Explanation:

Equation of the reaction:

CaO(s) + 2H+(aq) -----> Ca2+(aq) + H2O(g)

The ∆Hrxn would be for one mole of CaO reacted or 2 moles of H+, whichever is the limiting reactant.

Number of moles = mass ÷ molar mass

Molar mass of CaO = 40 + 16

= 56 g/mol

moles of CaO = 2.90/56

= 0.0518 mol

Number of moles = concentration × volume

moles of HCl = 400 × 10^-3 × 1.500 = 0.6 moles

Moles of HCl = moles of H+

From the equation, 1 mole of CaO reacted with 2 moles of H+ to give 1 mole of water.

To find the limiting reagent,

0.6 mole of H+/2 moles of H+ × 1 mole of CaO

= 0.3 moles of CaO(> 0.0518 moles)

So, CaO is limiting reactant.

∆H = m × Cp × ∆T

m = density × volume

= 400 × 1

= 400 g

Cp = 4.184 J/g-ºC

∆T = +6 ºC

∆H = 400 × 4.184 × 6

= 10041.6 J

Since the reaction is exothermic,

∆Hrxn = -∆H/mol(CaO)

= -10041.6/0.0518

= -193853 J

= -193.9 kJ/mol.

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