C 0.289 why but because of decimals i’m gonna go with that one
Answer:
CaCO3 (s) → CaO (s) + CO2 (g)
The mass of carbonate that must have reacted was 43.03 grams
Explanation:
CaCO3 → CaO + CO2
Relation between reactant and product is 1:1
Let's apply the Ideal Gas Law to find out the moles of CO2 which were produced.
P . V = n . R . T
1 atm . 23 L = n . 0.082 L.atm/mol.K . 653K
(1atm . 23L) / (0.082 mol.K/L.atm . 653K) = n
0.43 moles = n
0.43 moles of CO2, were produced from 0.43 moles of CaCO3.
Molar weight of CaCO3 = 100.08 g/m
Mass = Molar weight . moles
Mass = 100.08 g/m 0.43 m = 43.03 g
Answer:
42672 kJ
Explanation:
Chemical reaction:
C₂H₄ + H₂ → C₂H₆ + heat
Now we calculate the number of moles of ethane C₂H₆:
number of moles = mass / molecular weight
number of moles of ethane = 10.1 / 30 = 0.336 kmoles = 336 moles
And determine the heat released:
if 127 kJ are released when 1 mole of C₂H₆ is produced
then X kJ are released when 336 moles of C₂H₆ is produced
X = (127 × 336) / 1 = 42672 kJ
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