How many Silver Coins (weighing 26g each) would it take to provide 5 moles of Silver (Ag)?
5 moles Ag = 5*atomic mass Ag = ?? grams. But Ag coins are not pure Ag; in fact the percentage varies and some contain no Ag at all.
??grams Ag/percent(as a fraction) = ??grams coins.
Then ??g coins/26 = # coins
To determine the number of silver coins required to provide 5 moles of silver, we need to use the concept of molar mass.
1. First, let's determine the molar mass of silver (Ag). The molar mass of silver is equal to the atomic mass of silver, which is 107.87 grams per mole.
2. Calculate the mass of 5 moles of silver:
Mass = Number of moles × Molar mass
Mass = 5 moles × 107.87 g/mol
Mass = 539.35 grams
3. Since each silver coin weighs 26 grams, divide the mass (539.35 grams) by the weight of each coin (26 grams) to find the number of coins:
Number of coins = Mass / Weight of each coin
Number of coins = 539.35 g / 26 g ≈ 20.75
Therefore, it would take approximately 20.75 silver coins to provide 5 moles of silver. However, since you cannot have a fraction of a coin, you would need to round up to the nearest whole number. Thus, you would need a minimum of 21 silver coins.