4gm of lead nitrate reacts with 1.17g of sodium chloride and form 3gm of lead chloride how much odium nitrate will form?
"Moles (NaCl)=\\frac{1.17}{58.44}=0.02moles"
Ratio of "NaCl" and "NaNO_3" is 1:1
Hence
"Moles (NaNO_3)=0.02moles." 4gm
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