The number of moles of NalO3 present in the deposit is 83.23 mol NaIO3.
The mass of NaNO3 deposit present is 2000. Ib Molar mass of NaNO3 is 85.03 g/mol .
Calculate the number of moles of NaNO3
present in the 2000 pounds of NaNO3.
2000 lb x (453.592 gX 1 mol NaNO3)/ 1 lbX 85.03 g
=1.067x10^4 mol NaNO3
The percentage of NaIO3 present in the NaNO3 deposit is 0.78 mol%.
Multiply the number of moles of NaNO3 with the mole percentage of the NaIO3, to get the actual number of moles of NaIO3 present in the deposit.
0.78% (1.067×10* mol NaNO3)=0.78 mol NaIO3/100 mol NaNO,
= 83.23 mol
Thus, the number of moles of NalO3 present in the deposit is 83.23 mol NaIO3
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