17.37grams
Given the chemical reaction between ammonia and sulfuric acid expressed as:
[tex]2NH_3+H_2SO_4\rightarrow2NH_4+SO_4[/tex]Given the following parameter
Mass of H2SO4 = 50 grams
Determine the moles of sulfuric acid
[tex]\begin{gathered} moles=\frac{mass}{molar\text{ mass}} \\ moles\text{ of }H_2SO_4=\frac{50g}{98.079gmol^{-1}} \\ moles\text{ }of\text{ }H_2SO_4=0.5098mole \end{gathered}[/tex]According to stoichiometry, 2moles of ammonia reacted with 1 mole of sulfuric acid, hence the number of moles of ammonia needed is given as:
[tex]\begin{gathered} moles\text{ }of\text{ }NH_3=\frac{2}{1}\times0.5098 \\ moles\text{ }of\text{ }NH_3=1.020moles \end{gathered}[/tex]Determine the required mass of NH3
[tex]\begin{gathered} Mass\text{ of NH}_3=moles\times molar\text{ mass} \\ Mass\text{ of NH}_3=1.020moles\times\frac{17.031g}{mol} \\ Mass\text{ of NH}_3=17.37grams \end{gathered}[/tex]Therefore the mass of ammonia (NH3) needed to completely react with 50 grams of sulfuric acid is 17.37grams
