Answer to Question #143722 in General Chemistry for james

Question

How many atoms of oxygen are present in 211 kg of lead nitrate?

Answer 1

Solution:

molar mass of Pb(NO₃)₂ = 331.2 g/mol

1 mol of Pb(NO₃)₂ contains 6 moles of O

Number of particles N = Avogadro constant N_A × amount n

N of O = 211000g Pb(NO₃)₂ / 311.2g/mol Pb(NO₃)₂ × 6mol O × 6.02×10²³ = 2.3×10²⁷ atoms O

Answer: there are 2.3*10²⁷ atoms of O in 211 kg of lead nitrate