5.00 g of (Zn) was added to 0.800 moles of nitrate ion in acidic media.
Calculate the volume of (NO) gas collected at STP.
Zn(s) + No3^-(aq) ---------> ZN^2+(aq) + NO(g)
(Must balance the reaction first).
Thank you.
Correction:
Zn(s) + NO3^-(aq) ---------> ZN^2+(aq) + NO(g)
3Zn(s) + 2NO3^-(aq) + 8H^+ ---------> 3Zn^2+(aq) + 2NO(g) + 4H2O
This is a limiting reagent (LR) problem. Which is the LR?
mols Zn = grams/atomic mass = approx 0.08
Convert mols Zn to mols NO. That's approx 0.08 x (2 mols NO/3 mols Zn) = about 0.05 which is less than 0.8 mols from the NO3^-.
Then use PV = nRT and solve for V in liters.
To calculate the volume of NO gas collected at STP, we first need to balance the equation for the reaction between Zn and NO3-.
The balanced equation is:
3Zn(s) + 2NO3-(aq) + 8H+(aq) -> 3Zn^2+(aq) + 2NO(g) + 4H2O(l)
From the balanced equation, we can see that 3 moles of Zn react with 2 moles of NO3-. Therefore, the ratio of Zn to NO is 3:2.
Given:
Mass of Zn = 5.00 g
Molar mass of Zn = 65.38 g/mol
First, let's calculate the moles of Zn:
Moles of Zn = mass of Zn / molar mass of Zn
Moles of Zn = 5.00 g / 65.38 g/mol
Moles of Zn = 0.076 mol
Since the ratio of Zn to NO is 3:2, we can calculate the moles of NO produced:
Moles of NO = (2/3) × moles of Zn
Moles of NO = (2/3) × 0.076 mol
Moles of NO = 0.051 mol
Now that we have the moles of NO, we can calculate the volume of gas collected at STP. STP (Standard Temperature and Pressure) conditions are defined as 0 degrees Celsius (273.15 K) and 1 atmosphere (atm) of pressure. At STP, 1 mole of any ideal gas occupies 22.4 L.
Volume of NO gas = Moles of NO × 22.4 L/mol
Volume of NO gas = 0.051 mol × 22.4 L/mol
Volume of NO gas = 1.1424 L
Therefore, the volume of NO gas collected at STP is approximately 1.1424 L.