Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
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Answer is: the molar mass od sodium carbonate (Na₂CO₃) is 106.0 g/mol.
M(Na₂CO₃) = 2 · Ar(Na) + Ar(C) + 3 · Ar(O).
M(Na₂CO₃) = 2 · 23 + 12 + 3 · 16 · g/mol.
M(Na₂CO₃) = 46 + 12 + 48 · g/mol.
M(Na₂CO₃) = 106 g/mol; molar mass of sodium carbonate.
Ar is relative atomic mass (the ratio of the average mass of atoms of a chemical element to one unified atomic mass unit) of an element.
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Convalent Bond energy is low
