Question #79474 Chemistry Stoichiometry Stoichiometry 1 Answer anor277 Oct 6, 2016 A bit under #5*g# of ammonia will be produced. Explanation: #(NH_4)_2SO_4(aq) + Ca(OH)_2(aq) rarr 2NH_3(aq) + CaSO_4(s)darr + 2H_2O# #"Moles of "Ca(OH)_2# #=# #(10.0*g)/(74.09*g*mol^-1)=0.135*mol# And thus, given the stoichiometry, #0.270*mol# of ammonia will be generated. This has a mass of #0.270*molxx17.03*g*mol^-1=??g# Answer link Related questions How do you solve a stoichiometry problem? What is stoichiometry? Question #93ac6 Why do we study stoichiometry? How many grams of NaOH is produced from #1.20 x 10^2# grams of #Na_2O#? #Na_2O + H_2O -> 2NaOH# How many grams of Na2O are required to produce 1.60 x 102 grams of NaOH? Na2O + H2O ---> 2 NaOH What mass of iron is needed to react with 16.0 grams of sulfur? 8 Fe + S8 ---> 8 FeS According to 8 Fe + S8 ---> 8 FeS How many grams of FeS are produced? 12.00 moles of NaClO3 will produce how many grams of O2? 2 NaClO3 ---> 2 NaCl + 3 O2 How many grams of NaCl are produced when 80.0 grams of O2 are produced? 2 NaClO3 ---> 2 NaCl + 3 O2 See all questions in Stoichiometry Impact of this question 1634 views around the world You can reuse this answer Creative Commons License