The mass of sodium chloride, NaCl produced from the reaction is 18474 g
MgCl₂ + 2Na —> 2NaCl + Mg
Molar mass of MgCl₂ = 24 + (35.5 × 2) = 95 g/mole
Mass of MgCl₂ from the balanced equation = 1 × 95 = 95 g = 0.095 Kg
Molar mass of Na = 23 g/mole
Mass of Na from the balanced equation = 2 × 23 = 46 g = 0.046 Kg
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mole
Mass of NaCl from the balanced equation = 2 × 58.5 = 117 g = 0.117 Kg
SUMMARY
From the balanced equation above,
0.095 Kg of MgCl₂ reacted with 0.046 Kg of Na to produce 0.117 Kg of NaCl
From the balanced equation above,
0.095 Kg of MgCl₂ reacted with 0.046 Kg of Na
Therefore,
15 Kg of MgCl₂ will react with = (15 × 0.046) / 0.095 = 7.26 Kg of Na
From the above calculation, only 7.26 Kg of Na out of 50 Kg given is needed to react completely with 15 Kg of MgCl₂.
Thus, MgCl₂ is the limiting reactant
From the balanced equation above,
0.095 Kg of MgCl₂ reacted to produce 0.117 Kg of NaCl
Therefore,
15 Kg of MgCl₂ will react to produce = (15 × 0.117) / 0.095 = 18.474 Kg = 18474 g of NaCl
Thus, 18474 g of NaCl were obtained from the reaction
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