Question #119306

An experiment was carried out at 25 °C by dissolving an unknown mass of limestone granules into excess 1 mol dm–3 hydrochloric acid. The volume of gas produced over a period of 6 minutes was measured using a gas syringe. The result was plotted as shown in Figure 1.


to calculate the moles of gas produced at STP

[1 mole of any gas at STP has a volume of 22 400 cm3.]












(1 mark )

to calculate the mass of limestone granules used.

[Relative molecular mass of CaCO3 is 100.]

Expert's answer

m(CaCO3)M(CaCO3)=V(gas)22400{m(CaCO3) \over M(CaCO3)}= {V(gas) \over 22400}

m(CaCO3)=M(CaCO3)∗V(gas)22400m(CaCO3) = {M(CaCO3)*V(gas) \over 22400}

m=100∗4922400m = {100*49 \over 22400}

m = 0,22 g


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