Given the equation: CuCl2 + 2 NaNO3-- 2 Cu(NO3)2 + 2 NaCl, what is the limiting reagent if 15g of CuCl2 react with 20g of NaNO3?

Question

Given the equation: CuCl2 + 2 NaNO3--> 2 Cu(NO3)2 + 2 NaCl, what is the limiting reagent if 15g of CuCl2 react with 20g of NaNO3?

Serenity Allen · Accepted Answer

#CuCl_2 + 2NaNO_3 rarr Cu(NO_3)_2 + 2NaCl#

Moles of sodium nitrate, #(20.0*g)/(85.00*g*mol^-1)#;

moles of cupric chloride, #(15.0*g)/(134.35*g*mol^-1)#.

Which reagent, considering the stoichiometry, is in excess? How many moles of each reactant?

Luna Chen · Answer

#"CuCl"_2("aq") + "2NaNO"_3("aq")"##rarr#No Reaction

If this is a double replacement reaction or salt metathesis, then I have taken the liberty of including the symbol representing an aqueous solution, #("aq")#, to the formulas in the equation written as #"CuCl"_2("aq") + "2NaNO"_3("aq")"##rarr##"Cu(NO"_3)_2("aq") + "2NaCl(aq")#. In order for a double replacement reaction to occur, one of the products must be a solid precipitate (a salt), an insoluble gas, or water. Since all species are in aqueous solution, none of these products occur, which means there is no reaction. The result is an aqueous solution containing #"Cu"^(2+)"#, #"Cl"^(-)"#, #"Na"^(+)"#, and #"NO"_3""^(-)"# ions. The equation should be written as #"CuCl"_2("aq") + "2NaNO"_3("aq")"##rarr#No Reaction

Owen Ambrose · Answer

The #"CuCl"_2"# is the limiting reagent.

I'll omit the aqueous symbols if all you want to practice is finding the limiting reagent. Equilibrium Formula #"CuCl"_2 + "2NaNO"_3##rarr##"Cu(NO"_3)_2 + "2NaCl"# In order to determine limiting reagent, you must include one of the products. I'm going to choose #"NaCl"#. We need to calculate the yield of #"NaCl"# with both reagents. First, ascertain the reagents' and sodium chloride's molar masses. Molar Masses #"CuCl"_2":##"134.452 g/mol"# #"NaNO"_3":##"84.994669 g/mol"# #"NaCl":##"58.442769 g/mol"# The limiting reagent is the one that produces the least amount of NaCl after dividing the given mass by its molar mass, multiplying by the mol ratio between NaCl and the reagent from the balanced equation, and then multiplying by the molar mass of NaCl. #15cancel"g CuCl"_2xx(1cancel"mol CuCl"_2)/(134.452cancel"g CuCl"_2)xx(2cancel"mol NaCl")/(1cancel"mol CuCl"_2)xx(58.442769"g NaCl")/(1cancel"mol NaCl")="13.0 g NaCl"# #20cancel"g NaNO"_3xx(1cancel"mol NaNO"_3)/(84.994669cancel"g NaNO"_3)xx(2cancel"mol NaCl")/(2cancel"mol NaNO"_3)xx(58.442769"g NaCl")/(1cancel"mol NaCl")="13.8 g NaCl"# The #"CuCl"_2"# is the limiting reagent. Visit this website for more details: https://tutor.hix.ai

Cooper Adams · Answer

undefined To find the limiting reagent, we need to calculate the number of moles of each reactant. Then, we compare the stoichiometric coefficients in the balanced equation to determine which reactant is limiting.

First, calculate the molar mass of each substance:
- CuCl2: 63.5 (Cu) + 2 * 35.5 (Cl) = 134 g/mol
- NaNO3: 23 (Na) + 14 (N) + 3 * 16 (O) = 85 g/mol

Now, calculate the number of moles of each reactant:
- Moles of CuCl2 = 15 g / 134 g/mol = 0.112 moles
- Moles of NaNO3 = 20 g / 85 g/mol = 0.235 moles

According to the balanced equation, the stoichiometric ratio between CuCl2 and NaNO3 is 1:2. Therefore, 0.112 moles of CuCl2 would require 0.224 moles of NaNO3 to react completely. Since we have more moles of NaNO3 available (0.235 moles), CuCl2 is the limiting reagent in this reaction.

HIX Tutor · Answer

undefined To find the limiting reagent, follow these steps:

**Step 1: Calculate the moles of CuCl2 and NaNO3.**

1. **Molar mass of CuCl2:**
   - Cu = 63.55 g/mol
   - Cl = 35.45 g/mol (2 Cl = 70.90 g/mol)
   - Total = 63.55 + 70.90 = 134.45 g/mol

2. **Moles of CuCl2:**
   $$
   \text{Moles of CuCl2} = \frac{15 \, \text{g}}{134.45 \, \text{g/mol}} \approx 0.111 \, \text{mol}
   $$

3. **Molar mass of NaNO3:**
   - Na = 22.99 g/mol
   - N = 14.01 g/mol
   - O = 16.00 g/mol (3 O = 48.00 g/mol)
   - Total = 22.99 + 14.01 + 48.00 = 85.00 g/mol

4. **Moles of NaNO3:**
   $$
   \text{Moles of NaNO3} = \frac{20 \, \text{g}}{85.00 \, \text{g/mol}} \approx 0.235 \, \text{mol}
   $$

**Step 2: Use the balanced equation to find the ratio needed.**

From the equation: 
- 1 mole of CuCl2 reacts with 2 moles of NaNO3.

**Step 3: Calculate the required moles of NaNO3 for the available moles of CuCl2.**
$$
\text{Moles of NaNO3 needed} = 0.111 \, \text{mol CuCl2} \times 2 = 0.222 \, \text{mol NaNO3}
$$

**Step 4: Compare available moles of NaNO3 with moles needed.**

- You have 0.235 mol of NaNO3, but you need 0.222 mol.
- NaNO3 is enough.

**Step 5: Check for CuCl2.**

You have 0.111 mol of CuCl2 and it needs 0.111 mol of CuCl2 for 0.222 mol of NaNO3.

**Step 6: Conclusion**

Since CuCl2 is less than needed, CuCl2 is the limiting reagent.

**Final Answer:** CuCl2 is the limiting reagent.

Given the equation: CuCl2 + 2 NaNO3--&gt; 2 Cu(NO3)2 + 2 NaCl, what is the limiting reagent if 15g of CuCl2 react with 20g of NaNO3?

Given the equation: CuCl2 + 2 NaNO3--> 2 Cu(NO3)2 + 2 NaCl, what is the limiting reagent if 15g of CuCl2 react with 20g of NaNO3?