Determine the number of moles of the compound and determine the number of each type of atom in each of
the following
(a) 2.12 g of potassium bromide, kbr
(b) 0.1488 g of phosphoric acid, h3po4
(c) 23 kg of calcium carbonate, caco3
(d) 78.452 g of aluminium sulfate, al2(so4)3
0.1250 mg of caffeine, c8h10n4o2

Explanation:

Moles is denoted by given mass divided by the molecular mass ,  

Hence ,  

n = w / m

n = moles ,  

w = given mass ,  

m = molecular mass .

From the question ,  

( a ) potassium bromide, KBr

As we know , the molecular mass of KBr = 119 g/mol

from the question ,

given mass of KBr = 2.12 g

using , the above formula to calculate the moles ,

n = w / m  = 2.12 g / 119 g/mol = 0.0178 mol.

The moles of the corresponding atoms are -

moles of K = 0.0178 mol

moles of Br = 0.0178 mol.

(b) phosphoric acid, H₃PO₄

As we know , the molecular mass of H₃PO₄ = 98 g/mol

from the question ,

given mass of  H₃PO₄ = 0.1488 g

using , the above formula to calculate the moles ,

n = w / m  = 0.1488 g  / 98 g/mol = 0.0015 mol.

The moles of the corresponding atoms are -

moles of H = 3 * 0.0015 = 0.0045 mol

moles of P = 0.0015 mol

moles of 0 = 4* 0.0015 = 0.0060 mol

(c) calcium carbonate, CaCO₃

As we know , the molecular mass of CaCO₃ = 100 g/mol

from the question ,

given mass of  CaCO₃ =  23 kg

Since , 1 kg = 1000 g

given mass of CaCO₃ =  23 * 1000 = 23000 g

using , the above formula to calculate the moles ,

n = w / m  = 23000 g  / 100 g/mol = 230 mol .

The moles of the corresponding atoms are -

moles of Ca = 230 mol

moles of C = 230 mol

moles of O = 3 * 230 = 690 mol

(d) aluminium sulfate, Al₂(SO₄)₃

As we know , the molecular mass of Al₂(SO₄)₃ = 342 g/mol

from the question ,

given mass of Al₂(SO₄)₃ = 78.452 g

using , the above formula to calculate the moles ,

n = w / m  = 78.452 g / 342 g/mol = 0.229 mol.

The moles of the corresponding atoms are -

moles of Al = 2 * 0.229 = 0.458 mol

moles of S = 3 * 0.229 = 0.687 mol

moles of O = 12 * 0.229 = 2.748 mol.

(e) Caffeine, C₈H₁₀N₄O₂

As we know , the molecular mass of C₈H₁₀N₄O₂ = 194 g/mol

from the question ,

given mass of C₈H₁₀N₄O₂ = 0.1250 mg

Since , 1 mg = 1 / 1000 g

given mass of C₈H₁₀N₄O₂ = 0.1250 / 1000 g = 0.000125 g

using , the above formula to calculate the moles ,

n = w / m  = 0.000125 g  / 194  g/mol = 6.44 * 10 ⁻⁷ mol.

The moles of the corresponding atoms are -

moles of C = 8 * 6.44 * 10 ⁻⁷ mol. = 51.52  * 10 ⁻⁷ mol.

moles of H = 10 * 6.44 * 10 ⁻⁷ mol. = 64.4  * 10 ⁻⁷ mol.

moles of N = 4 * 6.44 * 10 ⁻⁷ mol. = 25.76  * 10 ⁻⁷ mol.

moles of O = 2 * 6.44 * 10 ⁻⁷ mol. = 12.88  * 10 ⁻⁷ mol.

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

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