What volume of dry NO(g) at STP could be produced by reacting 8.74 g of Cu with an excess of HNO3?
3Cu + 8 HNO3→ 3 Cu(NO3)2 + 2 NO(g) + 4 H2O(l)
mols Cu = grams/atomic mass = 8.75/63.5 = ?
mols NO produced = mols Cu x (2 mols NO/3 mols Cu) = ?
Then you know 1 mol NO occupies a volume of 22.4 L @ stp
Thank you, this really helped
Well, first of all, let's appreciate the fact that Cu and HNO3 got together and decided to have a chemical party. How considerate of them!
Now, let's crunch some numbers to find out the volume of the "NO(g) guest" at this party. We need to determine the number of moles of NO produced from the given mass of Cu.
The molar mass of Cu is 63.55 g/mol, so using some scientific math magic, we find that 8.74 g of Cu is equal to 0.138 mol.
According to the balanced equation, 3 moles of Cu will produce 2 moles of NO. Therefore, 0.138 mol of Cu will produce (2/3) * 0.138 mol of NO, which is approximately 0.092 mol of NO gas.
Now, since we're at STP (Standard Temperature and Pressure), we can assume that 1 mole of any ideal gas occupies 22.4 liters. Therefore, 0.092 mol of NO gas will occupy approximately 2.06 liters.
So, after all that math, the volume of dry NO(g) produced at STP by reacting 8.74 g of Cu with an excess of HNO3 is approximately 2.06 liters. Now that's quite the gas volume for a chemical party!
To determine the volume of dry NO(g) produced at STP, we need to use the given mass of Cu and the balanced chemical equation.
Step 1: Calculate the moles of Cu
The molar mass of Cu is 63.55 g/mol.
Moles of Cu = Mass of Cu / Molar mass of Cu
Moles of Cu = 8.74 g / 63.55 g/mol
Step 2: Determine the limiting reactant
Since there is an excess of HNO3, Cu is the limiting reactant and will determine the amount of NO(g) produced.
Step 3: Calculate the moles of NO(g)
From the balanced equation, we can see that the mole ratio between Cu and NO(g) is 3:2.
Moles of NO(g) = Moles of Cu × (2 moles of NO(g) / 3 moles of Cu)
Moles of NO(g) = (8.74 g / 63.55 g/mol) × (2/3)
Step 4: Convert moles of NO(g) to volume at STP
At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters.
Volume of NO(g) = Moles of NO(g) × Molar volume at STP
Volume of NO(g) = (8.74 g / 63.55 g/mol) × (2/3) × 22.4 L/mol
Perform the calculations to obtain the final answer.
To find the volume of dry NO(g) at STP produced by reacting 8.74 g of Cu with an excess of HNO3, we need to use the stoichiometry of the balanced equation.
First, we need to determine the number of moles of Cu. To do this, we can use the molar mass of copper (Cu), which is 63.55 g/mol. We can calculate the moles of Cu using the formula:
moles of Cu = mass of Cu / molar mass of Cu
moles of Cu = 8.74 g / 63.55 g/mol
Next, we use the stoichiometry of the balanced equation to find the moles of NO(g) produced. According to the balanced equation:
3 moles of Cu = 2 moles of NO
So, the moles of NO can be calculated as:
moles of NO = (moles of Cu) x (2 moles of NO / 3 moles of Cu)
Now that we have the moles of NO, we can use the ideal gas law to find the volume of NO(g) at STP. The ideal gas law equation is:
PV = nRT
Where:
P is the pressure (at STP, it is 1 atm)
V is the volume
n is the number of moles
R is the ideal gas constant (0.0821 L.atm/mol.K)
T is the temperature (at STP, it is 273 K)
Since we want to find the volume, we rearrange the equation:
V = (nRT) / P
Now, we can substitute the values into the equation:
V = (moles of NO) x (R) x (T) / P
V = (moles of NO) x (0.0821 L.atm/mol.K) x (273 K) / 1 atm
Finally, we can calculate the volume of dry NO(g) at STP:
V = (moles of NO) x (22.414 L/mol)
Keep in mind that at STP, 1 mole of any gas occupies 22.414 L.
By following this procedure, you can calculate the volume of dry NO(g) at STP produced by reacting 8.74 g of Cu with an excess of HNO3.