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35.0 grams of lithium carbonate is dissolved in water and excess calcium nitrate is added. The precipitate is collected, dried and weighed. How much should the dry precipitate weigh?

User Biwek
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1 Answer

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Step-by-step explanation:

Lithium carbonate is dissolved in water and calcium nitrate is added in excess. That would be a double replacement reaction, these compounds will exchange their ions and lithium nitrate and calcium carbonate will be produced. The balanced equation of the reaction is:

Li₂CO₃ (aq) + Ca(NO₃)₂ (aq) ----> CaCO₃ (s) + 2 LiNO₃ (aq)

The precipitate that is formed is calcium carbonate, which is insoluble. We have to determine the amount of it produced. First we have to convert the mass of lithium carbonate into moles using its molar mass.

mass of Li₂CO₃ = 35.0 g

molar mass of Li₂CO₃ = 73.9 g/mol

moles of Li₂CO₃ = 35.0 g * 1 mol/(73.9 g)

moles of Li₂CO₃ = 0.474 moles

Li₂CO₃ (aq) + Ca(NO₃)₂ (aq) ----> CaCO₃ (s) + 2 LiNO₃ (aq)

According to the balanced equation, 1 mol of Li₂CO₃ will produce 1 mol of CaCO₃ (when reacting with excess calcium nitrate). So the molar ratio between them is 1 to 1. We can use that relationship to find the number of moles of CaCO₃ that are produced by 0.474 moles of Li₂CO₃.

1 mol of Li₂CO₃ : 1 mol of CaCO₃ molar ratio

moles of CaCO₃ = 0.474 moles of Li₂CO₃ * 1 mol of CaCO₃/(1 mol of Li₂CO₃)

moles of CaCO₃ = 0.474 moles

And finally we can convert back to grams the 0.474 moles of CaCO₃ using its molar mass.

molar mass of CaCO₃ = 100.1 g/mol

mass of CaCO₃ = 0.474 moles * 100.1 g/(1 mol)

mass of CaCO₃ = 47.4 g

Answer: the dry precipitate should weigh 47.4 g

User Tigran Iskandaryan
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