Respuesta :
Answer: 218.4 g of [tex]CaO[/tex] will be produced from 3.9 moles of [tex]CaCO_3[/tex]
Explanation:
The balanced chemical equation is:
[tex]CaCO_3\rightarrow CaO+CO_2[/tex]
According to stoichiometry :
1 mole of [tex]CaCO_3[/tex] produce = 1 mole of [tex]CaO[/tex]
Thus 3.9 moles of [tex]CaCO_3[/tex] will produce=[tex]\frac{1}{1}\times 3.9=3.9moles[/tex] of [tex]CaO[/tex]
Mass of [tex]CaO=moles\times {\text {Molar mass}}=3.9moles\times 56g/mol=218.4g[/tex]
Thus 218.4 g of [tex]CaO[/tex] will be produced from 3.9 moles of [tex]CaCO_3[/tex]
218.4 grams of CaO is produced using 3.9 moles CaCO₃.
How we calculate weight of any substance from moles?
Moles of any substance will be define as:
n = W / M
Given chemical reaction is:
CaCO₃ → CaO + CO₂
From the above equation it is clear that according to the concept of stoichiometry 1 mole of CaCO₃ is producing 1 mole of CaO. By using above formula we calculate grams as follow:
W = n × M, where
n = no. of moles of CaO = 3.9 moles
M = molar mass of CaO = 56 g/mole
W = 3.9 × 56 = 218.4 g
Hence, 218.4 grams of CaO is produced.
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https://brainly.com/question/13314627
