Respuesta :
Answer:
81.732KJ
Explanation:
Reaction : CaCO₃(s) → CaO(s) + CO₂(g), +556KJ
⇒To decompose 1 mole of CaCO₃, 556KJ of energy is required
The molecular weight of CaCO₃ is 40 + 12 + 3×16
⇒ molecular weight of CaCO₃ = 100 g
∴ 100 g of CaCO₃ requires 556KJ of energy
Need to find out how much energy is required by 14.7 g of CaCO₃
⇒ [tex]\frac{100}{14.7}[/tex] = [tex]\frac{556}{E}[/tex]
⇒ E = [tex]\frac{556}{100}[/tex]×14.7 KJ = 81.732 KJ
∴ Energy required to decompose 14.7 g of CaCO₃ is 81.732 KJ
Answer:
81.732 kilo Joules of energy is required to decompose 14.7 grams of calcium carbonate.
Explanation:
Moles of calcium carbonate = [tex]\frac{14.7 g}{100 g/mol}=0.147 mol[/tex]
[tex]CaCO_3(s) + 556 kJ\rightarrow CaO(s) + CO_2(g)[/tex]
According to reaction, 1 mole of calcium carbonate requires 556 kilo Joules of energy to decompose:
Then 0.147 moles of calcium carbonate will need:
[tex]0.147 \times 556 kJ=81.732 kJ[/tex]
81.732 kilo Joules of energy is required to decompose 14.7 grams of calcium carbonate.
