10.0 g H2O reacts with 4.5 g Na to produce NaOH and H2. Identify the limiting reactant. *
2Na+2H2O→2NaOH+H22Na+2H_2O\to 2NaOH+H_22Na+2H2O→2NaOH+H2
moles of sodium=4.5/23=0.1957moles4.5/23=0.1957moles4.5/23=0.1957moles
moles of water = 10/18=0.5556moles10/18=0.5556moles10/18=0.5556moles
Moles of sodium are less than moles of water hence it is the limiting reactant.
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