how many moles of Lead Nitrate are required to produce 80 grams of oxygen
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1
Answer:
32g of oxygen =1mole
Hence 48 g =( 1x48)/32= 1.5 moles
2Pb(NO3)2 -------2 PbO + 4NO2+O2
According to this equation 2 moles of lead nitrate gives 1 mole of oxygen
1mole of oxygen = 2 moles of lead nitrate
Hence 1.5 moles of oxygen = 2×1.5 = 3 moles of lead nitrate
Explanation:
Answered by
0
Answer:
3 mole
Explanation:
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