A different rigid 5.00L cylinder contains 0.176mol of NO(g) at 298 k. A 0.176mol sample of 02(g) is added to the cylinder, where a reaction occurs to produce NO2(g)

(d) write the balanced equation for the reaction.
(e) calculate the total pressure, in atm, in the cylinder at 298k after the reation is complete

To write the balanced equation for the reaction, we need to determine the products formed when NO(g) and O2(g) react to produce NO2(g).

The reaction between NO and O2 can be represented by the following balanced equation:

2NO(g) + O2(g) → 2NO2(g)

This means that for every 2 moles of NO(g) and 1 mole of O2(g), 2 moles of NO2(g) are produced.

Now let's move on to calculating the total pressure in the cylinder after the reaction is complete.

First, let's determine the initial number of moles of NO2(g) in the cylinder. Since 0.176 moles of NO(g) react to produce 0.176 moles of NO2(g), the final number of moles of NO2(g) in the cylinder is also 0.176 moles.

Next, we need to apply the ideal gas law equation to calculate the total pressure. The ideal gas law equation is:

PV = nRT

where 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.

In this case, we are given the volume of the cylinder (5.00 L) and the temperature (298 K), and we have calculated the number of moles of NO2(g) (0.176 moles).

Plugging these values into the ideal gas law equation, we have:

P * 5.00 L = 0.176 mol * 0.0821 L·atm/(mol·K) * 298 K

Simplifying the equation, we get:

P = (0.176 mol * 0.0821 L·atm/(mol·K) * 298 K) / 5.00 L

Calculating the right side of the equation:

P = 2.4633 atm

Therefore, the total pressure in the cylinder at 298 K, after the reaction is complete, is approximately 2.4633 atm.