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Balancing chemical equations involves ensuring that the same number of each type of atom is present on both sides of a chemical equation. This is crucial to accurately represent chemical reactions. In the given equation, \( \mathrm{Fe}_2\mathrm{O}_3(s) + \mathrm{CO}(g) \rightarrow \mathrm{Fe}(s) + \mathrm{CO}_2(g) \), we start by balancing the iron (Fe) atoms.
1. We have 2 Fe atoms in \( \mathrm{Fe}_2\mathrm{O}_3 \), so we need 2 Fe atoms on the product side. We place a coefficient of 2 in front of \( \mathrm{Fe} \).2. Next, we balance oxygen (O). There are 3 oxygen atoms in \( \mathrm{Fe}_2\mathrm{O}_3 \), and each \( \mathrm{CO}_2 \) has 2, so using 3 \( \mathrm{CO}_2 \) molecules gives us 6 O atoms on the right. Adjusting the CO reactant to 3 balances the carbon as well. Therefore, the balanced equation becomes:\[ \mathrm{Fe}_2\mathrm{O}_3(s) + 3\mathrm{CO}(g) \rightarrow 2\mathrm{Fe}(s) + 3\mathrm{CO}_2(g) \]Remember, balancing equations ensures the conservation of atoms, reflecting their immutability during chemical reactions.

Stoichiometry is the scientific study of the quantities of reactants and products in a chemical reaction. It is rooted in the balanced chemical equation.
First, let's convert 0.350 kg of \( \mathrm{Fe}_2\mathrm{O}_3 \) to grams (350 g), simplifying calculations. Using its molar mass (159.69 g/mol), we find:\[ \text{moles of } \mathrm{Fe}_2\mathrm{O}_3 = \frac{350\ g}{159.69\ g/mol} = 2.191\ mol \]Using the balanced equation, the mole ratio instructs stoichiometric proportions:- For CO: \( 2.191\ mol \ \mathrm{Fe}_2\mathrm{O}_3 \times \frac{3\ mol\ \mathrm{CO}}{1\ mol\ \mathrm{Fe}_2\mathrm{O}_3} = 6.573\ mol\ \mathrm{CO} \)- Convert to mass: \( 6.573\ mol\ \mathrm{CO} \times 28.01\ g/mol = 184\ g \ \mathrm{CO} \)This confirms the precise amount of reactants and products through required conversions.
For products:- Fe: \( 2.191\ mol\ \mathrm{Fe}_2\mathrm{O}_3 \times \frac{2\ mol\ \mathrm{Fe}}{1\ mol\ \mathrm{Fe}_2\mathrm{O}_3} = 4.382\ mol\ \mathrm{Fe} \)- Convert to grams: \( 4.382\ mol\ \mathrm{Fe} \times 55.845\ g/mol = 245\ g \ \mathrm{Fe} \)And:- CO\(_2\): \( 2.191\ mol\ \mathrm{Fe}_2\mathrm{O}_3 \times \frac{3\ mol\ \mathrm{CO}_2}{1\ mol\ \mathrm{Fe}_2\mathrm{O}_3} = 6.573\ mol\ \mathrm{CO}_2 \)- Convert to grams: \( 6.573\ mol\ \mathrm{CO}_2 \times 44.01\ g/mol = 289\ g \ \mathrm{CO}_2 \)These stoichiometric calculations underscore how precise mole ratios enable predictability for reaction outcomes.

The law of conservation of mass dictates that in a closed system, mass is neither created nor destroyed by chemical reactions. Instead, the mass of reactants equals the mass of products, a fundamental principle in chemistry.
In this reaction:- Mass of reactants = 350 g \( \mathrm{Fe}_2\mathrm{O}_3 \) + 184 g \( \mathrm{CO} \) = 534 g- Mass of products = 245 g \( \mathrm{Fe} \) + 289 g \( \mathrm{CO}_2 \) = 534 gBoth sides of the equation total to 534 grams, confirming that every atom from the reactants is present in the products.
This verification showcases the principle that total mass remains constant over a chemical reaction's course. It's imperative for calculations in chemistry, ensuring that predictions accurately reflect reality. By adhering to this principle, chemists consistently achieve balance through careful measurement and adjustments during experiments.