g Aluminum sulfite reacts with sodium hydroxide to form sodium sulfite and aluminum hydroxide according to the following equation. Al2(SO3)3 6 NaOH> 3 Na2SO3 2 Al(OH)3a)Determine the limiting reactant if 18.0 g of Al2(SO3)3reacts with 16.0 g of NaOH. b)Determine the number of moles and mass of Al(OH)3produced. c)Determine the number of molesand mass of Na2SO3produced. d)Determine the grams of excess reagent left over in the reaction.

ill try to . no promises tho.

explanation:

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.