Ammonium nitrate
Ammonium = NH4+
Nitrate = NO3-
<span>Na</span>₂<span>CO</span>₃<span> + 2 HCl = 2 NaCl + CO</span>₂<span> + H</span>₂<span>O
106 g Na</span>₂CO₃ -------- 2 x 36.5 g HCl
<span> x g Na</span>₂CO₃ ---------- 750 g HCl
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Mass of Na</span>₂CO₃ = 750 x 106 / 2 x 36.5
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Mass ( Na</span>₂CO₃) = 79500 / 73
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Mass = 1089.04 g
Number of moles = 1089.04 / 106 => 10.27 mols of Na</span>₂CO₃
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hope this helps!
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The enthalpies of formation of each of the compound involved in the chemical reaction presented above are given below:
CO2: -393.5 kJ/mol
CO: -99 kJ/mol
O2: 0 kJ/mol
As observed O2 will not have enthalpy of formation as it is a pure substance.
To calculate for the enthalpy of reaction,
enthalpy of formation of products - enthalpy of formation of reactants
= (-99 kJ/mol) - (-393.5 kJ/mol)
= 294.5 kJ/mol
ANSWER: 294.5 kJ/mol
C distillation you would evaporate the water into another container leaving you with the salt free water