Question

What volume of nitrogen dioxide (in L) is produced from the complete reaction of 16.87g of lead(II)nitrate at STP 2Pb(NO3)2 (g) -->2PbO(s) + 4NO2(g) + O2

Answer 1

mPb(NO₃)₂: 331 g/mol
----------------------------

2Pb(NO₃)₂ ----> 2PbO + 4NO₂ + O₂
2×331g 4×22,4dm³

662g Pb(NO₃)₂ ---------- 89,6 dm³ NO₂
16,87g Pb(NO₃)₂ ------- X
X = (89,6×16,87)/662
X = 2,2833 dm³ = 2,2833 L NO₂