Molar mass of NaHCO3 = 23 + 1 + 12 + 48 = 84 g/mol
Moles = 1/84 = 0.0119 mol
Molar mass of H3C6H5O7 = 192
Moles = 1/192 = 0.0052
To find limiting reagent, moles are divided by coefficient
0.0119 /3 = 0.00396
0.0052/1 = 0.0052
As 0.00396 < 0.0052
So, NaHCO3 is limiting reagent
Do rate if you understand
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