Question

How many grams of ammonia (NH3) are needed to completely react with 50 grams of sulfuric acid (H2SO4?)Round to two decimal places2 NH3 + H2SO4 + 2NH4+ SO4

Answer 1

17.37grams

Explanations:

Given the chemical reaction between ammonia and sulfuric acid expressed as:

`2NH_3+H_2SO_4\rightarrow2NH_4+SO_4`

Given the following parameter

Mass of H2SO4 = 50 grams

Determine the moles of sulfuric acid

`\begin{gathered} moles=\frac{mass}{molar\text{ mass}} \\ moles\text{ of }H_2SO_4=(50g)/(98.079gmol^(-1)) \\ moles\text{ }of\text{ }H_2SO_4=0.5098mole \end{gathered}`

According to stoichiometry, 2moles of ammonia reacted with 1 mole of sulfuric acid, hence the number of moles of ammonia needed is given as:

`\begin{gathered} moles\text{ }of\text{ }NH_3=(2)/(1)*0.5098 \\ moles\text{ }of\text{ }NH_3=1.020moles \end{gathered}`

Determine the required mass of NH3

`\begin{gathered} Mass\text{ of NH}_3=moles* molar\text{ mass} \\ Mass\text{ of NH}_3=1.020moles*(17.031g)/(mol) \\ Mass\text{ of NH}_3=17.37grams \end{gathered}`

Therefore the mass of ammonia (NH3) needed to completely react with 50 grams of sulfuric acid is 17.37grams