What are the partial pressures for all gases in 4NH3+O2-->4NO+6H2O that remained at the end, if the reaction took place in a sealed 50.0L tank and the final temperature is 345K?

You haven't provided enough information. You have no Keq and you have no starting amounts.

To find the partial pressures for all gases in the given reaction, we need to calculate the moles of each gas present at the end and then use the ideal gas law to determine the partial pressure. Let's go step by step:

Step 1: Balance the equation:
4NH3 + 5O2 --> 4NO + 6H2O

Step 2: Calculate the moles of each gas:
Given that the reaction took place in a sealed 50.0 L tank, we can assume that the final volume of the gases is also 50.0 L at the given temperature, 345 K.

Using the balanced equation, we can determine the stoichiometry of the reaction:
4 moles of NH3: 4 moles of NO
5 moles of O2: 4 moles of NO

From the stoichiometry, we can deduce that the number of moles of O2 present at the end is 5/4 times that of NO. Since the reaction produced 4 moles of NO, the moles of O2 at the end are (5/4) * 4 = 5 moles.

Similarly, the number of moles of NH3 present at the end is the same as that of NO, which is 4 moles.

Finally, the number of moles of H2O produced at the end is 6 times that of NO, so it is 6 * 4 = 24 moles.

Step 3: Use the ideal gas law to calculate the partial pressures:
The ideal gas law is given by: PV = nRT

For each gas, the pressure (P) is equal to the moles (n) times the gas constant (R) times the temperature (T), divided by the volume (V).

P(NH3) = (moles of NH3 * R * T) / V
P(O2) = (moles of O2 * R * T) / V
P(NO) = (moles of NO * R * T) / V
P(H2O) = (moles of H2O * R * T) / V

Step 4: Plug in the values and calculate the partial pressures:
Given:
R = 0.0821 L·atm/(mol·K)
T = 345 K
V = 50.0 L

P(NH3) = (4 * 0.0821 * 345) / 50.0
P(O2) = (5 * 0.0821 * 345) / 50.0
P(NO) = (4 * 0.0821 * 345) / 50.0
P(H2O) = (24 * 0.0821 * 345) / 50.0

Calculating these values will give you the partial pressures for each gas in the reaction.