Key Idea : The reagent which is present in smaller quantity is called the limiting reagent and the moles of product depends on it and number of moles = molecular weight weight PbO207.2+16=223.22HCl⟶2(35.5+1)=73PbCl2+H2O207.2+71=278.2 Here, 1 mole of PbO reacts with 2 moles of HCl, thus PbO is the limiting reagent ∵223.2 g PbO gives PbCl2=278.2 g ∴6.5 gPbO will give PbCl2 =223.2278.2×6.5 g=223.2×278.2278.2×6.5 mol=0.029 mol
How many moles of lead (II) chloride will be formed from a reaction between 6.5 ~g of PbO and 3.2 ~g of HCl ?
NEET UG 2008 — Chemistry Organic Chemistry
2008mcqmedium
How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl ?
Official previous-year question
Held on 30 Apr 2008 · Verified 9 Jul 2026.
Options
- A
0.044
- B
0.333
- C
0.011
- D
0.029
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