Our balanced equation is Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag.
The coefficients of reactants tell us that 1 mole of Cu reacts per 2 moles of AgNO₃. We want to know the mass of Cu that will react with a mass (specifically, 420 g) of AgNO₃. To start, we must convert the mass of AgNO₃ to moles of AgNO₃:
Moles of AgNO₃ = Mass of AgNO₃ ÷ Molar Mass of AgNO₃
Moles of AgNO₃ = 420 g AgNO₃ ÷ 169.88 g AgNO₃/mol AgNO₃
Moles of AgNO₃ = 2.47 moles of AgNO₃.
Since only half as many moles of Cu is consumed in reacting with the AgNO₃, the number of moles of Cu that will react with 2.47 moles of AgNO₃ would be half of 2.47, or 1.24 moles of Cu.
Finally, we can convert the molar quantity of Cu to mass by multiplying the number of moles of Cu by the molar mass of Cu:
Mass of Cu = Moles of Cu × Molar Mass of Cu
Mass of Cu = 1.24 moles Cu × 63.546 g Cu/moles Cu
Mass of Cu = 78.6 g of Cu
So, 78.6 grams of Cu are needed to react with 420 g of AgNO₃.
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The way to solve the second question is identical to how we solved the first question. The coefficient of Ag is 2, so the molar ratio between the Ag produced and Cu reacted is 2:1. In other words, half as many moles of Ag produced is equivalent to the number of moles of Cu that was consumed in the reaction.
So, as we did with the first question, we first convert the mass of the Ag produced to moles. The molar mass of Ag is 107.8682 g/mol. Dividing 98.5 g of Ag by the molar mass gives us 0.913 moles of Ag produced. The number of moles of Cu that reacted to produce 0.913 moles of Ag would be 0.913/2 = 0.457 moles of Cu.
Lastly, multiplying 0.457 moles of Cu by the molar mass of Cu, we obtain the mass of Cu (in grams) that reacted: 0.457 × 63.546 = 29.0 g of Cu.
So, 29.0 grams of Cu reacted to produce 98.5 grams of Ag.
Note: I have provided the answers for both questions to three significant figures; while 420 technically has only two significant figures, it would seem consistent to treat it as having three since 98.5 has three digits and three significant figures.
