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Fer the Free Respense you may ase a colculeter an 1 Pcrlodle TaNe. produciog calcium chlorids, water, and carbon dioxide. (s) Wrise a balancel equation for the rastion. \[ \begin{array}{l} \left.\left.\mathrm{Ca}_{2} \mathrm{O}_{3}(\Omega)+\mathrm{HCl}_{(s)}\right) \longrightarrow \mathrm{CaCl}_{2}(4)\right)+\mathrm{H}_{2} \mathrm{OCO}+\mathrm{CO}_{2}(\mathrm{~s}) \\ \text { Aas: } \mathrm{CaCO}_{3}(\theta)+2 \mathrm{HCI}(\mathrm{O}) \longrightarrow \mathrm{CaCl}_{2}(4 \rho)+\mathrm{H}_{2} \mathrm{O}(\rho)+\mathrm{CO} \text { (s } \end{array} \] (b) When the rcaction aas cornglets, 1.55 g of carbcn diovide gas was collocted. Hrer many moles of calrium cubceate were consemed ia the resction? (c) If all Cocalcium carbocate initially presect in the kmple was curnamed ia the reaction, what pervent by cuss of the sass ple was \&us ss eakines carbocate? (d) If be inert enterial was only silicon dioxile, what sas the mole fncticn of silicon diovide is the mikture? \[ \text { mole fructicn }=N_{6}=\frac{n_{1}}{n_{2 n}} \] or have eo etteat Juatify your sesponse.

Ask by Johnson Floyd. in Papua New Guinea
Mar 14,2025

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(a) The balanced equation is: \[ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \] (b) 0.0352 moles of calcium carbonate were consumed. (c) The percent yield of calcium carbonate is 100%. (d) The mole fraction of silicon dioxide in the mixture cannot be determined without knowing the amount of \( \text{SiO}_2 \) present.

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To balance the equation for the given reaction, we recognize that limestone (calcium carbonate, CaCO₃) reacts with hydrochloric acid (HCl) to produce calcium chloride (CaCl₂), water (H₂O), and carbon dioxide (CO₂). The balanced equation would look like this: \[ \text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g) \] Now, onto the fun part! When considering 1.55 g of carbon dioxide collected, we can find moles using the molar mass of CO₂ (approximately 44.01 g/mol). So, dividing 1.55 g by 44.01 g/mol gives us about 0.035 moles of CO₂. Since the reaction shows a 1:1 ratio of CaCO₃ to CO₂, we also consumed about 0.035 moles of calcium carbonate. Next, if we assume all CaCO₃ was consumed and started with 1.55 g of it, the percentage consumed can be calculated as (moles consumed / initial moles) * 100%. Just a quick engaging nugget! The world of chemistry is like a magic show; reactants turn into products, and the show might just collect some gas along the way! Finally, let's theorize about silicon dioxide in an inert mixture with respect to mole fraction. If silicon dioxide (SiO₂) is the only inert material, we'll enter the realm of the ideal gas law. If we know the total moles in the mixture, we can calculate the mole fraction using \( X_{SiO2} = \frac{n_{SiO2}}{n_{Total}} \). It's just a fun little fraction game where the total pie is shared—only in chemical style!

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