Pearson Edexcel IAL Chemistry 1.7 Reacting mass calculationsPractise using balanced equations, molar masses and experimental masses to find products, excess reagents and mixture composition.SyllabusFirst assessment 2019CourseChemistry YCH11LevelAS
Exam pointscalculate maximum product mass from a limiting reactant and a balanced equationcompare reactant mole amounts to state which reagent is in excessuse mass change or precipitate mass to find composition or solution concentration
1.7—Chemical equations to calculate reacting masses and vice versa question 1[Maximum number: 1]This question is about magnesium, magnesium oxide and magnesium sulfate.Hydrated crystals of magnesium sulfate, MgSO4⋅7H2O\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}MgSO4⋅7H2O, can be made by reacting magnesium with sulfuric acid.In an experiment, magnesium was added to 30.0 cm330.0 \mathrm{~cm}^{3}30.0 cm3 of 0.500 moldm−30.500 \mathrm{~mol} \mathrm{dm}^{-3}0.500 moldm−3 sulfuric acid.[Mr\left[M_{\mathrm{r}}\right.[Mr value: MgSO4⋅7H2O=246.4Ar\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}=246.4 \quad A_{\mathrm{r}}MgSO4⋅7H2O=246.4Ar value: Mg=24.3]\left.\mathrm{Mg}=24.3\right]Mg=24.3]Calculate the mass of magnesium needed to react with the sulfuric acid.Show AnswerMark as masteredmass of MgExample of calculation0.015/1.5×10−2×24.3=0.3645( g)0.015 / 1.5 \times 10^{-2} \times 24.3=0.3645(\mathrm{~g})0.015/1.5×10−2×24.3=0.3645( g)Ignore SF except 1SFTE on (e)(i).(1)Add to Test