Home › WAEC WASSCE › Chemistry › Stoichiometry of Chemical Reactions › A 0.4550 g mixture treated with excess Ba(NO3)2 …
A 0.4550 g mixture treated with excess Ba(NO3)2 gives 0.6168 g of BaSO4 (233.43 g/mol). What is the mass percent of MgSO4 (120.37 g/mol)?
A26.5%
B51.2%
C138%
D69.9%
Answer & Solution
Correct answer: D. 69.9%
1. Convert the precipitate to moles: 0.6168 g divided by 233.43 g/mol gives 0.0026421 mol BaSO4.
2. The equation pairs 1 mol MgSO4 with 1 mol BaSO4, so the same molar amount of MgSO4 was present.
3. Convert back to mass: 0.0026421 mol times 120.37 g/mol gives 0.3181 g MgSO4.
4. Mass percent equals the analyte mass divided by the sample mass, times 100.
5. That is 0.3181 divided by 0.4550, which is 0.699.
6. Multiplying by 100 gives about 69.9% MgSO4 in the mixture.
7. Dividing the sample mass by the precipitate mass instead gives 138%, an impossible percentage.
_Source: OpenStax Chemistry (CC BY 4.0), Ch 4 "Stoichiometry of Chemical Reactions", section 4.5 Quantitative Chemical Analysis_
Related questions
In the reaction between sodium and chlorine, sodium is oxidised, which means chlorine:Ordering the three ways of writing a reaction from most to least detail about ions gives:When one element appears in more than one formula on the same side of an equation, you musWhen balancing a combustion equation, an odd number of oxygen atoms on the product side isCompounds that produce hydroxide ions by chemically reacting with water, rather than by diA double arrow is used in an acid equation to indicate that the acid is:A neutralisation reaction has an acid and a base as reactants and typically produces:Sodium hydroxide is classified as a strong base because in water it: