Answer:
![m_(NH_3)=4.25kgNH_3](https://img.qammunity.org/2021/formulas/chemistry/college/cr28dliap79tn67nkux2tlanwkehyfiufq.png)
Step-by-step explanation:
Hello,
In this case, for the production of ammonium nitrate we shall consider the following chemical reaction:
![NH_3+HNO_3\rightarrow NH_4NO_3](https://img.qammunity.org/2021/formulas/chemistry/college/ovm9saqn4l4woi7ezody1yk4we5y39rx2b.png)
Hence, since the molar mass of ammonium nitrate is 80 g/mol and the molar mass of ammonia is 17 g/mol, we could compute the required mass of ammonia to produce 20 kg of ammonium nitrate by using kilo-based units:
![m_(NH_3)=20kgNH_4NO_3*(1kmol)/(80kgNH_4NO_3)*(1kmolNH_3)/(1kmolNH_4NO_3)*(17kgNH_3)/(1kmolNH_3) \\\\m_(NH_3)=4.25kgNH_3](https://img.qammunity.org/2021/formulas/chemistry/college/d3rnxnv5xfcjuibrvvd2r43dijyd808ldi.png)
Best regards.