If you start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?

Md Ariful Islam

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Md Ariful Islam

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Pb(NO3)2 + 2NaI = 2NaNO3 + PbI2

n = m / M

M (Pb(NO3)2) = 331.2 g/mol

M (NaI) = 150 g/mol

M (NaNO3) = 85 g/mol

n (Pb(NO3)2) = 25 / 331.2 = 0.08 mol

n (NaI) = 15 / 150 = 0.1 mol

According to the equation, n (NaI) = 2 x n (Pb(NO3)2). From the given reactant quantities, NaI is the limiting reactant. n (NaNO3) = n (NaI). The amount of NaNO3 to be formed is:

m (NaNO3) = n x M = 0.1 x 85 = 8.5 g

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