1. When solid copper metal reacts with nitric acid, the products include copper (II) ion and nitrogen monoxide gas. If 11.99 g of solid copper are reacted with 153 mL of 0.322 M HNO3, how many liters of nitrogen monoxide will be produced?

2. When solid copper metal reacts with nitric acid, the products include copper (II) ion and nitrogen monoxide gas. If 11.99 g of solid copper are reacted with 153 mL of 0.322 M HNO3, how many liters of nitrogen monoxide will be produced? (Assume the released gas is at standard temperature and pressure.)

3Cu + 8HNO3 ==> 3Cu(NO3)2 + 2NO + 4H2O

mols Cu = grams/molar mass
Using the coefficients in the balanced equation, convert mols Cu to mols NO.
Now convert mols NO to L at STP. 1 mol NO gas at STP occupies 22.4 L.

1. Oh, copper and nitric acid, sounds like a fascinating chemistry experiment! Let's crunch some numbers, shall we?

First things first, we need to find the number of moles of copper involved. To do that, we can use the molar mass of copper (Cu) which is approximately 63.55 g/mol. So, 11.99 g of copper would be:

11.99 g / 63.55 g/mol ≈ 0.1889 mol of copper.

Now, let's focus on the nitric acid (HNO3) part. We have 153 mL, but we need to convert that to liters because chemistry loves consistency. There are 1000 mL in a liter, so:

153 mL / 1000 mL/L = 0.153 L of HNO3.

Converting to moles is a piece of cake! We multiply the volume (L) by the molarity (mol/L):

0.153 L × 0.322 mol/L ≈ 0.0493 mol of HNO3.

According to the balanced equation, the ratio of copper to nitrogen monoxide (NO) is 1:1. So, the number of moles of NO produced will be the same as the number of moles of copper, which is 0.1889 mol.

Now, to answer your question, we need to convert this amount from moles to liters. At standard temperature and pressure (STP), 1 mole of any gas occupies approximately 22.4 liters. Therefore:

0.1889 mol × 22.4 L/mol ≈ 4.2296 L of nitrogen monoxide.

So, approximately 4.23 liters of nitrogen monoxide gas will be produced. Enjoy the chemical circus!

2. Ah, the same chemical circus act with solid copper and nitric acid! This time, we'll snap our fingers and presto, assume the gas is at standard temperature and pressure (STP).

We've already calculated that 0.1889 mol of nitrogen monoxide (NO) will be produced. Using the molar volume at STP (22.4 L/mol), we can convert this mol value to liters:

0.1889 mol × 22.4 L/mol ≈ 4.2296 L of nitrogen monoxide.

So, at STP, approximately 4.23 liters of nitrogen monoxide gas will be produced. Tada! The clown chemistry show never disappoints.

To answer both questions, we need to calculate the stoichiometry of the reaction and then use the given amounts to determine the quantity of the desired product.

1. The balanced equation for the reaction between copper and nitric acid is:
Cu + 4HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O

From the balanced equation, we can see that for every 1 mole of Cu reacted, 2 moles of NO are produced. We need to first convert the given mass of Cu to moles:

Molar mass of Cu = 63.55 g/mol
Number of moles of Cu = mass / molar mass = 11.99 g / 63.55 g/mol = 0.1888 mol

Since the stoichiometric ratio between Cu and NO is 1:2, there will be twice the number of moles of NO produced:

Number of moles of NO produced = 2 * 0.1888 mol = 0.3776 mol

Next, we need to convert the moles of NO to volume using the ideal gas law:

PV = nRT

At standard temperature and pressure (STP), the volume of 1 mole of any ideal gas is 22.4 L. Therefore:

Volume of NO produced = number of moles of NO * 22.4 L/mol = 0.3776 mol * 22.4 L/mol = 8.46 L

Therefore, 8.46 liters of nitrogen monoxide gas will be produced.

2. Since the released gas is assumed to be at standard temperature and pressure (STP), we can directly use the molar volume of gases at STP, which is 22.4 L/mol.

Therefore, the volume of nitrogen monoxide gas produced will be 8.46 L, assuming the reaction occurs under standard temperature and pressure conditions.

To solve both of these questions, we need to use stoichiometry, which is the calculation of quantities in a chemical reaction. Here's how we can solve them step by step:

1. To find the number of moles of copper (Cu), we need to divide the given mass of copper by its molar mass. The molar mass of copper is 63.55 g/mol.

Number of moles of Cu = 11.99 g / 63.55 g/mol = 0.1887 mol

2. The balanced chemical equation for the reaction between copper and nitric acid is:

3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

From the balanced equation, we can see that for every 3 moles of copper, 2 moles of nitrogen monoxide (NO) are produced.

Therefore, the number of moles of NO produced can be calculated using the stoichiometric ratio:

Number of moles of NO = (0.1887 mol Cu) * (2 mol NO / 3 mol Cu) = 0.1258 mol NO

3. Now, let's calculate the volume of nitrogen monoxide gas using the ideal gas law, which states:

PV = nRT

P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

As given in the second question, we assume the released gas is at standard temperature (273 K) and pressure (1 atm). Therefore, we can rearrange the ideal gas law equation to solve for volume:

V = (nRT) / P

Plug in the values:

V = (0.1258 mol) * (0.0821 L·atm/mol·K) * (273 K) / (1 atm) ≈ 2.581 L

Therefore, approximately 2.581 liters of nitrogen monoxide gas will be produced.

So, the answer to both questions is 2.581 liters of nitrogen monoxide.