Find the volume of gas that is released when 26 grams of Na2CO3 interacts with HCL.
| November 19, 2020 | education
Na2CO3 (1 mol) + 2HCl = 2NaCl + CO2 (1 mol) + H2O
M (Na2CO3) = 23 * 2 + 12 + 16 * 3 = 106g / mol
n (Na2CO3) = 26/106 = (approximately) 0.25 mol
According to the reaction equation:
n (Na2CO3) refers to n (CO2) as 1/1
which means: n (CO2) = 0.25 mol
V = Vm * n
Vm = 22.4 l / mol
V (CO2) = 0.25 * 22.4 = 5.6L
Answer: 5.6 l.
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