2 The reaction between zinc carbonate and hydrochloric acid can be used to make zinc chloride. The equation for this reaction is: ZnCO_(3)(s)+2HCl(aq)arrow ZnCl_(2)(aq)+H_(2)O(l)+CO_(2)(g) 6.25 g of ZnCO_(3) was added to a solution containing 1.825 g of HCI. a Which reactant is in excess? Show your reasoning. b What mass of zinc chloride would you expect to make? (A, values: Zn=65,Cl=35.5,H=1)

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2 The reaction between zinc carbonate and hydrochloric acid can be used to make zinc chloride. The equation for this reaction is: ZnCO_(3)(s)+2HCl(aq)arrow ZnCl_(2)(aq)+H_(2)O(l)+CO_(2)(g) 6.25 g of ZnCO_(3) was added to a solution containing 1.825 g of HCI. a Which reactant is in excess? Show your reasoning. b What mass of zinc chloride would you expect to make? (A, values: Zn=65,Cl=35.5,H=1)

Answer 1

1. \(\mathrm{ZnCO}_{3}\) is in excess. 2. The mass of \(\mathrm{ZnCl}_{2}\) produced is \(3.4 \, \mathrm{g}\).

Answer 2

## Step 1: Write the balanced chemical equation: \[\mathrm{ZnCO}_{3}(\mathrm{~s}) + 2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{ZnCl}_{2}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) + \mathrm{CO}_{2}(\mathrm{~g})\] ## Step 2: Calculate the molar masses of the reactants and products: ### \(\mathrm{ZnCO}_{3}\): \[\mathrm{ZnCO}_{3} = 65 (\mathrm{Zn}) + 12 (\mathrm{C}) + 16 \times 3 (\mathrm{O}) = 125 \, \mathrm{g/mol}\] ### \(\mathrm{HCl}\): \[\mathrm{HCl} = 1 (\mathrm{H}) + 35.5 (\mathrm{Cl}) = 36.5 \, \mathrm{g/mol}\] ### \(\mathrm{ZnCl}_{2}\): \[\mathrm{ZnCl}_{2} = 65 (\mathrm{Zn}) + 35.5 \times 2 (\mathrm{Cl}) = 136 \, \mathrm{g/mol}\] ## Step 3: Calculate the number of moles of each reactant: ### Moles of \(\mathrm{ZnCO}_{3}\): \[\frac{6.25 \, \mathrm{g}}{125 \, \mathrm{g/mol}} = 0.05 \, \mathrm{moles}\] ### Moles of \(\mathrm{HCl}\): \[\frac{1.825 \, \mathrm{g}}{36.5 \, \mathrm{g/mol}} = 0.05 \, \mathrm{moles}\] ## Step 4: Determine the limiting reactant: ### According to the balanced equation, 1 mole of \(\mathrm{ZnCO}_{3}\) reacts with 2 moles of \(\mathrm{HCl}\). Therefore, 0.05 moles of \(\mathrm{ZnCO}_{3}\) would require: \[0.05 \, \mathrm{moles} \times 2 = 0.10 \, \mathrm{moles \, of \, HCl}\] Since we only have 0.05 moles of \(\mathrm{HCl}\), \(\mathrm{HCl}\) is the limiting reactant. ## Step 5: Calculate the mass of \(\mathrm{ZnCl}_{2}\) produced: ### Moles of \(\mathrm{ZnCl}_{2}\) produced: \[\text{Since } \mathrm{HCl} \text{ is the limiting reactant, the moles of } \mathrm{ZnCl}_{2} \text{ produced will be } \frac{0.05}{2} = 0.025 \, \mathrm{moles}\] ### Mass of \(\mathrm{ZnCl}_{2}\): \[0.025 \, \mathrm{moles} \times 136 \, \mathrm{g/mol} = 3.4 \, \mathrm{g}\]

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