How do I answer this?

Question

Copper reacts with nitric acid in the following unbalanced equation:

___ Cu (s) + ___ HNO₃ (aq) --> ___ CuNO₃(aq) + ___ NO(g) + ___ H₂O(l)

How many milliliters of 0.155 M HNO₃ is needed to produce 155.25 milligrams of cuprous nitrate?

Answer 1

Hi Alexis,

The first thing you want to do is balance this equation. I like to write the equation the same way you have it, with the spaces next to each compound, and then make a list underneath of all the elements and how many of each there are. You want to do this for both sides of the equation.

___ Cu (s) + ___ HNO₃ (aq) --> ___ CuNO₃(aq) + ___ NO(g) + ___ H₂O(l)

Cu: 1 Cu: 1

H: 1 H: 2

N: 1 N: 2

O: 3 O: 5

This can be a little confusing because you have to add up the total number of N, for example, even though there is 1 N in CuNO₃ and 1 N in NO.

Now, you want to add coefficients to the different compounds to try and match the number of each element. Make sure you keep track of how many total of each element there are as you add coefficients. For example, if you add a 2 in front of CuNO₃, you will then have 3 N, with 2 from this compound and 1 for NO. This is a little difficult to draw out in this format, but here are my final coefficients. I encourage you to do this on your own and make sure you can get the same answer so that you know you are able to do it in the future on your own.

3 Cu (s) + 4 HNO₃ (aq) --> 3 CuNO₃(aq) + NO(g) + 2 H₂O(l)

This equation is on the more difficult side to balance. I would recommend starting with the N and adding H and O as applicable until you balance those. Because Cu is isolated on the left side, you don't have to worry about balancing that until the end. Remember that you should reduce the coefficients at the end to be the lowest values possible. For example, if you get 6/8/6/2/4 as your balancing coefficients, you can reduce them all by two to get the values I have proposed above.

Now, onto the actual question! In this equation, you will be working "backward" or from products to reactants. Finding the molecular weights for the compounds in question is a good first start.

CuNO₃ = 125.550 g/mol

HNO₃ = 62.012 g/mol

Now, make sure you take note of the units of the amounts given to you. I would convert the amount of cuprous nitrate to grams to make the calculations a little easier.

155.25 milligrams cuprous nitrate x (1 g cuprous nitrate / 1000 mg cuprous nitrate) = 0.15525 g cuprous nitrate.

Now, you can use the mole ratios to determine how many moles of HNO₃ you'd need. You should stop at moles of HNO₃ because that will make working with the molarity easiest.

0.15525 g CuNO₃ x (1 mol CuNO₃ / 125.550 g CuNO₃) x (4 mol HNO₃ / 3 mol CuNO₃) = 0.00165 mol HNO₃

Now, from knowing the amount of HNO₃ you will need, you can use the molarity of the solution you have to find the volume.

0.00165 mol HNO₃ x (1 L HNO₃ / 0.155 mol HNO₃) = 0.01064 L HNO₃

The question asks for milliliters, so just convert the answer in L to mL.

0.01064 L HNO₃ x (1000mL HNO₃ / 1L HNO₃) = 10.64 mL HNO₃