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How many grams of CaO can be produced using 3.9 moles CaCO3?1
CaCO3 --> 1CaO + 1CO *


Sagot :

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.

To know more about moles, visit the below link:

https://brainly.com/question/13314627