Use stoichiometry 3.75moles k* 39.10g k\1 mole k= 146.63g k
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Answer:
1.48×10²⁴ Cu atoms
Explanation:
For this question you need to use Avogadro's number 6.022×10²³atoms.
2.45 moles of Cu ×
= 1.47539×10²⁴ atoms.
The moles cancel out so you are left with atoms.
Since there are 3 significant figures in the question there should be 3 significant figures in your answer, which is 1.48×10²⁴ Cu atoms.
<u>Given:</u>
Mass of Ag = 1.67 g
Mass of Cl = 2.21 g
Heat evolved = 1.96 kJ
<u>To determine:</u>
The enthalpy of formation of AgCl(s)
<u>Explanation:</u>
The reaction is:
2Ag(s) + Cl2(g) → 2AgCl(s)
Calculate the moles of Ag and Cl from the given masses
Atomic mass of Ag = 108 g/mol
# moles of Ag = 1.67/108 = 0.0155 moles
Atomic mass of Cl = 35 g/mol
# moles of Cl = 2.21/35 = 0.0631 moles
Since moles of Ag << moles of Cl, silver is the limiting reagent.
Based on reaction stoichiometry: # moles of AgCl formed = 0.0155 moles
Enthalpy of formation of AgCl = 1.96 kJ/0.0155 moles = 126.5 kJ/mol
Ans: Formation enthalpy = 126.5 kJ/mol
