1) 3.0g of Ga2O3 was added to excess HCL to form GaCl3 and
water. After purification and drying, the mass of GaCl3 was found
to be 3.5g. Calculcate the percentage yield of the
experiment.
2) When 0.1357g of copper (II) sulfate hydrate (CuSO4.xH2O) is
heated in air, it loses its water of crystallisation to form
0.0867g of anyhydrous white copper sulfate crystals. Determine the
number of water molecules per mole of copper (II) sulfate
hydrate
3) 10cm^3 of a hydrocarbon, CxHy were exploded with 100cm^3 of
oxygen. The volume of the resulting gas mixture was 80cm^3. When
the product was treated with aqueous sodium hydroxide, the
resulting volume was 50cm^3. All gaseous volumes were measured
under room temperatures and pressure. What is the molecular formula
of the hydrocarbon?
4) A 2.00g of limestone is allowed to react with 100cm^3 of 0.200
mol/dm^3 H2SO4. The excess acid required 12.40cm^3 of 0.100
mol/dm^3 NaOH for neutralisation. Calculate the percentage of
calcium carbonate in the limestone
Q1. Write the balanced equation. From sample mass of reactant, find
moles of reactant, hence apply stoichiometry to find expected moles
of product, then convert to expected sample mass of product. Take
actual sample mass of product, divided by expected sample mass 3.0g
of product.
Q2. Find moles of hydrate CuSO4.xH2O in terms of x
(ie. sample mass / molar mass). From sample mass of anhydrate,
find moles of anhydrate. This is also the number of moles of the
hydrate. Equate the two moles (anhydrate as number, hydrate in
terms of x) to solve for x.
Q3. Write the balanced equation for combustion of hydrocarbon,
in terms of x and y. 80-50=30cm3 was CO2 (since 2OH- + CO2 -->
CO3 2- + H2O). 10cm3 of CxHy generated 30cm3 of CO2, hence x = 3.
At rtp, H2O is liquid. Hence the remaining 50cm3 must have been
excess O2. Hence you know how much O2 was used up (per 10cm3 of
CxHy). Hence, applying stoichiometry of the balanced
combustion equation, already knowing x, you can solve for y
(since the coefficient of O2 used in the balanced equation is
in terms of x and y).
Q4. For this qn, H2SO4 is considered a strong diprotic
acid. From moles of OH- used, you can find moles of H+ in excess.
Hence find moles of H+ which reacted (ie. take total moles of H+
from H2SO4, minus away moles of H+ in excess) with the CaCO3, hence
apply stoichiometry to calculate moles of CO3 2- (note
that CO3 2- is a diprotic base), which is the same as moles of
CaCO3. Then convert the moles of sample mass. Take this actual
sample mass, divided by actual sample mass 2.0g.