If 29.0g of potassium reacts with 66.5g of barium nitride, what mass of potassium nitride will be produced?

32.4 grams of K3N will be produced

Explanation:

Step 1: Data given

Mass of potassium = 29.0 grams

Mass of barium nitride = 66.5 grams

Atomic mass of potassium = 39.10 g/mol

Molar mass of barium nitride = 440 g/mol

Step 2: The balanced equation

6K + Ba3N2 → 2K3N + 3Ba

Step 3: Calculate moles

Moles = mass / molar mass

Moles K = 29.0 grams / 39.10 g/mol

Moles K = 0.742 moles

Moles Ba3N2 = 66.5 grams / 440 g/mol

Moles Ba3N2 = 0.151 moles

Step 4: Calculate the limiting reactant

For 6 moles K we need 1 mol Ba3N2 to produce 2 moles K3N and 3 moles Ba

K is the limiting reactant. It will completely be consumed (0.742 moles). Ba3N2 is in excess. There will react 0.742/ 6 = 0.124 moles

There will remain 0.151 - 0.124 = 0.027 moles

Step 5: Calculate moles K3N

For 6 moles K we need 1 mol Ba3N2 to produce 2 moles K3N and 3 moles Ba

For 0.742 moles K we'll have 0.742/3 = 0.247 moles K3N

Step 6: Calculate mass K3N

Mass K3N = moles K3N * molar mass K3N

Mass K3N = 0.247 moles * 131.3 g/mol

Mass K3N = 32.4 grams

32.4 grams of K3N will be produced

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

space exploration has led to the development and technological improvement of many everyday items. (if this answer is wrong i will find the right answer) : )