3. What is the Limiting Reactant in the following equation if you start with 3 moles of NH3 with 6 moles of O2? 4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O(g)

4. How many grams of NO can be made from the previous equation and quantities?

3. NH3 is limiting

4. 90 g NO

Explanation:

3. To find the limiting reactant, you have to convert from moles of reactant to moles of product. In this case, because we need to know how many grams of NO in the second question, we will use NO as our product. We use the ratios found in the chemical formula.

Because NH3 makes less moles of NO, it is the limiting reactant.

4. Now we need the molar mass to convert the moles of NO into grams of NO. The molar mass of NO is 30.01g/mol. This is the equation.

Because there are is only one significant figure in the original numbers, we round to one significant figure = 90 g NO.

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.