79.8 g; 1
a) Mass of Fe2O3
Molar mass of Fe2O3 = (2x55.845 +3x16.00) g Fe2O3 = 159.69 g Fe2O3
Mass of Fe2O3 = 0.500 mol Fe2O3 x (159.69 g Fe2O3)/1 mol Fe2O3)
= 79.8 g Fe2O3
b) Coefficient of CaO
The balanced equation is
1CaO + CO2 → CaCO3
The coefficient of CaO is 1.
Answer : The mass of [tex]Fe_2O_3[/tex] are, 79.845 grams
In the given reaction, the coefficient for calcium oxide is, 1
Explanation :
Part 1 : Given,
Moles of [tex]Fe_2O_3[/tex] = 0.500 mole
Molar mass of [tex]Fe_2O_3[/tex] = 159.69 g/mole
[tex]\text{Mass of }Fe_2O_3=\text{Moles of }Fe_2O_3\times \text{Molar mass of }Fe_2O_3[/tex]
[tex]\text{Mass of }Fe_2O_3=0.500mole\times 159.69g/mole=79.845g[/tex]
Therefore, the mass of [tex]Fe_2O_3[/tex] are, 79.845 grams
Part 2 :
The given balanced chemical reaction will be,
[tex]CaO+CO_2\rightarrow CaCO_3[/tex]
This reaction is a balanced reaction in which 1 mole of calcium oxide react with 1 mole of carbon dioxide to give 1 mole of calcium carbonate as a product.
Hence, the coefficient for calcium oxide is, 1
