When sodium reacted, the amount of 0.5 mol with H2O produced hydrogen with a volume

| September 11, 2021 | education

When sodium reacted, the amount of 0.5 mol with H2O produced hydrogen with a volume of 4.2 liters, calculate the yield of the reaction product

1) Na2CO3 + 2HCl = 2NaCl + H2O + CO2.
For K2CO3, a similar reaction takes place.
At 100% yield of carbon dioxide product, 20.56 liters should be formed: (85%: 100%) = 20.56: 0.85 = 24.19 liters, this amount of substance is 24.19 liters: 22.4 liters / mol = 1.08 mol.
The same amount of sodium carbonate (or sodium) should be taken.
By mass it will be
for Na2CO3 106 g / mol * 1.08 mol = 114.5 g.
K2CO3 138 g / mol * 1.08 mol = 149 g.
2) 2Na + 2H2O = 2NaOH + H2.
If sodium reacted in an amount of 0.5 mol, then 2 times less hydrogen should be released, 0.5: 2 = 0.25 mol, or 0.25 * 22.4 = 5.6 liters.
The practical product yield in our case is equal to:
(4.2: 5.6) * 100% = 75%.



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