In the reaction 3Cu(s) + 8HNO₃(aq) $\rightarrow$ 3Cu(NO₃) ₂(aq) + 4H₂O(l) + 2NO(g)
6.3 g of Cu and 100.0 mL of 1.0 M HNO₃ are combined. Which of the following statements is true?

Question

In the reaction 3Cu(s) + 8HNO₃(aq) $\rightarrow$3Cu(NO₃)₂(aq) + 4H₂O(l) + 2NO(g) 6.3 g of Cu and 100.0 mL of 1.0 M HNO₃ are combined. Which of the following statements is true?

Quizplus · Accepted Answer

First, we need to write and balance the equation: 
3Cu(s) + 8HNO3(aq) -> 3Cu(NO3)2(aq) + 4H2O(l) + 2NO(g)

Next, we need to find the moles of Cu and HNO3. 
Moles of Cu = 6.3 g / 63.55 g/mol = 0.099 mol 
Moles of HNO3 = 1.0 M * 0.100 L = 0.100 mol

Now, we can determine the limiting reagent. Using the balanced equation, we see that the Cu and HNO3 have a 3:8 mole ratio. Therefore, we need 3/8 as many moles of HNO3 as we have moles of Cu to have a complete reaction. 
(3/8) * 0.099 mol = 0.0371 mol

Since we have 0.100 mol HNO3 and only need 0.0371 mol, HNO3 is in excess and Cu is the limiting reagent.

Using the mole ratio from the balanced equation, we can calculate the moles of H2O produced. 
Moles of H2O = (4/3) * moles of Cu = (4/3) * 0.099 mol = 0.132 mol

Finally, we can convert the moles of water produced to grams: 
Mass of H2O = 0.132 mol * 18.015 g/mol = 2.38 g

Therefore, statement D is true: HNO3 is the limiting reagent, and 0.062 mol Cu does not react.

In the Reaction 3Cu(s) + 8HNO3(aq) →\rightarrow→ 3Cu(NO3)2(aq) + 4H2O(l) + 2NO(g) 6

In the Reaction 3Cu(s) + 8HNO₃(aq) $\rightarrow$ 3Cu(NO₃)₂(aq) + 4H₂O(l) + 2NO(g)
6