calculations for solvay(carbonating ammonical brine) (1 Viewer)

[charbar]

ok heres the question:
CO2 + NH3 + NaCl + H2O ----> NaHCO3 + NH4Cl

Calculate teh mass of carbon dioxide needed to react with 14g of ammonia.

Im confused about calculating the CO2 as a volume of gas or in grams.. teh answer i have is in grams. please help!!!

 

[beta-omega]

first, you can not use volume, because its a moles question, and you have not been given the conditions under which the above is proceeded, meaning that it has to be in grams, plus they do state, calculate the mass. Ok, i'll work the thing out...

n (NH3) = m/M = 14/17 = 0.824mol
then the number of CO2 needed according to your reaction is the same number of moles as the NH3, since n (CO2): n (NH3)=1:1, therefore, you need 0.824mols.
now, rearranging the equation n=m/M, you get that mass= mols.molecular mass, therefore,
mass of CO2 = (44)(0.824)=36.3 grams.

 

[Plebeian]

-Calculate the amount of moles of ammonia in 14g.
Using moles = mass / molar mass, there are (14/17.034) = 0.8219 moles

-The ratio of CO2 to NH3 is 1:1, so there 0.8219 moles of CO2 are needed.

-Since each mole of CO2 weighs 44.01g, 0.8219 moles weighs 36.17g

-If it asked for volume, use the fact that a mole of any gas is 24.79L at 25ºC and 100kPa,
so the volume would be 0.8219 * 24.79 = 20.37L