if 2.34 moles of Tin IV Chloride reacts with 4.36 moles of sodium, according to the following reaction, what is the maximum numb

1.09 moles of tin (Sn)

Explanation:

The reaction we are examining involves tin chloride (SnCl₂) reacting with sodium (Na) to yield tin (Sn) and sodium chloride (NaCl):

SnCl₄ + 4 Na → Sn + 4 NaCl

Considering the chemical equation, this results in:

If 1 mole of SnCl₂ reacts with 4 moles of Na,

then 2.34 moles of SnCl₂ will react with X moles of Na.

X = (2.34 × 4) / 1 = 9.36 moles of Na

So, 2.34 moles of SnCl₂ can react with 9.36 moles of Na, but we have only 4.36 moles of Na on hand. Therefore, Na is the limiting reactant. Based on this, we can conclude:

If 4 moles of Na yield 1 mole of Sn,

then 4.36 moles of Na will produce Y moles of Sn.

Y = (4.36 × 1) / 4 = 1.09 moles of Sn

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limiting reactant