The molar mass of NO₂ is 46.0 g/mol
The molar mass of Pb (NO₃)₂ is 331.2 g/mol
First there is a need to find the number of moles of NO₂ via the stoichiometry of reaction:
2Pb(NO₃)₂ → 2PbO (s) + 4NO₂ (g) + 02 (g)
2 × 331.2 g = 4 × 46.0 g
16.87 g = x (mass of NO₂)
mass of NO₂ = 16.87 × 4 × 46 / 2 × 331.2
mass of NO₂ = 3104.08 / 662.4
mass of NO₂ = 4.686 g of NO₂
Now the number of moles are:
1 mole NO₂ = 46.0 g
x moles of NO₂ = 4.686 g
4.686 × 1 / 46.0 = 0.101 moles of NO₂
1 mole = 22.4 L (at STP)
0.101 moles of NO₂ = 0.101 × 22.4 / 1
= 2.26 L
Answer:
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