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# In which case, purity of the substance is $100\%$ ?(A) $1{\text{ mole of CaC}}{{\text{O}}_3}{\text{ gave 11}}{\text{.2L C}}{{\text{O}}_2}{\text{ (at STP)}}$ (B) $1{\text{ mole of MgC}}{{\text{O}}_3}{\text{ gave 40}}{\text{.0g MgO}}$ (C) $1{\text{ mole of NaHC}}{{\text{O}}_3}{\text{ gave 4g }}{{\text{H}}_2}{\text{O }}$ (D) $1{\text{ mole of Ca(HC}}{{\text{O}}_3}{{\text{)}}_2}{\text{ gave 1 mole C}}{{\text{O}}_2}{\text{ }}$

Last updated date: 20th Mar 2023
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Assume that all the given compounds are $100\%$ pure. So, $1{\text{ mole of CaC}}{{\text{O}}_3}$ should give ${\text{22}}{\text{.4L C}}{{\text{O}}_2}{\text{ (at STP)}}$ . This is determined from the stoichiometric ratio of moles of reactant to moles of carbon dioxide. Write a chemical equation and find the ratio of moles of calcium carbonate to carbon dioxide. $1{\text{ mole of MgC}}{{\text{O}}_3}$ should give $40.0{\text{ g MgO}}$ . ${\text{22}}{\text{.4L C}}{{\text{O}}_2}{\text{ (at STP)}}$ . This is determined from the stoichiometric ratio of moles of reactant to moles of magnesium oxide. Write a chemical equation and find the ratio of moles of magnesium carbonate to magnesium oxide. $1{\text{ mole of NaHC}}{{\text{O}}_3}$ should give $9{\text{ g }}{{\text{H}}_2}{\text{O}}$ . This is determined from the stoichiometric ratio of moles of reactant to moles of water. Write a chemical equation and find the ratio of moles of $NaHC{O_3}$ to water. $1{\text{ mole of Ca(HC}}{{\text{O}}_3}{)_2}$ should give $2{\text{ moles C}}{{\text{O}}_2}$ . This is determined from the stoichiometric ratio of moles of reactant to moles of carbon dioxide. Write a chemical equation and find the ratio of moles of ${\text{Ca(HC}}{{\text{O}}_3}{)_2}$ to carbon dioxide.
Percentage purity is calculated by dividing the mass of the pure chemical by the total mass of the sample given and then multiplying the result by $100$ . The percent yield is defined as the ratio of the actual yield to the theoretical yield expressed in the percentage form.