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How many grams of tin, Sn, are required to form 0.260 g of silver, Ag, by the following reaction?

2 Ag+(aq) + Sn(s) → 2 Ag(s) + Sn2+(aq)



Molar masses: Ag = 107.9 g mol-1 Sn = 118.7 g mol-1



Mass of Sn required to produce 0.260 g of Ag

1 Answer

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Answer and Explanation:

To determine the mass of tin (Sn) required to produce 0.260 g of silver (Ag), we need to use the stoichiometry of the reaction and the molar masses of the elements involved.

From the balanced chemical equation:

2 Ag+(aq) + Sn(s) → 2 Ag(s) + Sn2+(aq)

We can see that the mole ratio between Ag and Sn is 2:1. This means that for every 2 moles of Ag, we need 1 mole of Sn.

First, we need to convert the mass of Ag (0.260 g) to moles using its molar mass:

Molar mass of Ag = 107.9 g/mol

Number of moles of Ag = Mass of Ag / Molar mass of Ag

Number of moles of Ag = 0.260 g / 107.9 g/mol

Now, using the mole ratio, we can determine the number of moles of Sn required:

Number of moles of Sn = Number of moles of Ag / 2

Finally, we can convert the number of moles of Sn to grams using its molar mass:

Molar mass of Sn = 118.7 g/mol

Mass of Sn = Number of moles of Sn * Molar mass of Sn

By substituting the values into the equation, we can find the mass of Sn required to produce 0.260 g of Ag.

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