Although Cu is not sufficiently active to react with acids, it can be dissolved by concentrated nitric acid, which functions as an oxidizing agent according to the following equation:
Cu(s) + 4 HNO₃(aq) → Cu(NO₃)₂(aq) + 2 NO₂(g) + 2 H₂O(l)
b. Is 35.0 g of HNO3 sufficient to dissolve 5.00 g of copper?

Question

Although Cu is not sufficiently active to react with acids, it can be dissolved by concentrated nitric acid, which functions as an oxidizing agent according to the following equation:

Cu(s) + 4 HNO₃(aq) → Cu(NO₃)₂(aq) + 2 NO₂(g) + 2 H₂O(l)

b. Is 35.0 g of HNO3 sufficient to dissolve 5.00 g of copper?

Accepted Answer

When this problem, we are told that silver fully dissolves in hot concentrated nitric acid, which acts as a oxidizing agent to form the water soluble silver nitrate and a brown gas. One mole of silver plus two moles of nitric acid react to form one mole of silver nitrate plus one mole of nitrogen dioxide gas plus one mole of water. We're asked to determine whether 40 g of nitric acid is enough to dissolve 30 g of silver. We are provided with four possible answers. First three say yes, 40 g of nitric acid is enough to dissolve 30 g of silver. A says we can dissolve around 34.2 g of silver. B around 68.5 g of silver and C around 80 g of silver cancer D states. No, 40 g of nitric acid is not enough to dissolve 30 g of silver since it can only dissolve around 20 g of silver. So this began then by determining with 40 g of nitric acid, how much silver could we dissolve? And then compare that to the 30 g that we were given. So we'll start with 40 g or nitric acid we need to convert it to moles. So we can compare it to silver. So we'll make use of the molar mass multiply by one mole of nitric acid and divide by 63. grams of nitric acid for grams of nitric acid in the numerator and denominator cancels, we then go to our balanced reaction equation which says that one mole of silver will react with two moles from nitric acid. So we'll multiply by one mole of silver and dissolve or sorry, divide by two moles of nitric acid for moles of nitric acid in the numerator and denominator cancels and then we will convert our moles of silver to mass silver using its molar mass. So we multiply by molar mass of silver which is 107. g and divide by one mole of silver. So our moles of silver and the numerator and denominator cancels this is equal to 34.2 grams of silver. So when we look at our possible answers, then we can dissolve 34.2 g. So yes, we do have enough. So that then means that we'll eliminate. Answer D which says no. So answer B says that we can dissolve around 68.5 g. So we can't dissolve that much and we can't dissolve 80 g either. So that makes B and C incorrect. So answer A is the correct answer. Yes. 40 g of nitric acid is enough to dissolve 30 g of silver and can dissolve up to around 34.2 g of silver. Thanks for watching. Hope this helped.

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Key Concepts

Stoichiometry

Molar Mass

Limiting Reactant