## General, Organic, and Biological Chemistry: Structures of Life (5th Edition)

a. The balanced chemical equation for this reaction is: $$2NH_3 + 5F_2 \longrightarrow N_2F_4 + 6 HF$$ b. In order to produce 4.00 moles of HF, we need 1.33 moles of $NH_3$ and 3.33 moles of $F_2$. c. There are 142 g of $F_2$ d. 10.4 g of dinitrogen tetrafluoride can be produced.
a. 1. Identify the chemical formulas for each reactant and product: Reactants: Ammonia ($NH_3$) and fluorine $(F_2)$ Products: Dinitrogen tetrafluoride $(N_2F_4)$ and hydrogen fluoride $(HF)$ 2. Write the unbalanced equation: $$NH_3 + F_2 \longrightarrow N_2F_4 + HF$$ Balance the number of nitrogens: $$2NH_3 + F_2 \longrightarrow N_2F_4 + HF$$ Balance the number of hydrogens: $$2NH_3 + F_2 \longrightarrow N_2F_4 + 6HF$$ Finally, balance the number of fluorines: $$2NH_3 + 5F_2 \longrightarrow N_2F_4 + 6 HF$$ b. - Using the balance coefficents as conversion factors: $$4.00 \space moles \space HF \times \frac{ 2 \space moles \ NH_3 }{ 6 \space moles \space HF } = 1.33 \space moles \space NH_3$$ $$4.00 \space moles \space HF \times \frac{ 5 \space moles \ F_2 }{ 6 \space moles \space HF } = 3.33 \space moles \space F_2$$ c. $NH_3$ : ( 1.008 $\times$ 3 )+ ( 14.01 $\times$ 1 )= 17.03 g/mol $$\frac{1 \space mole \space NH_3 }{ 17.03 \space g \space NH_3 } \space and \space \frac{ 17.03 \space g \space NH_3 }{1 \space mole \space NH_3 }$$ $F_2$ : ( 19.00 $\times$ 2 )= 38.00 g/mol $$\frac{1 \space mole \space F_2 }{ 38.00 \space g \space F_2 } \space and \space \frac{ 38.00 \space g \space F_2 }{1 \space mole \space F_2 }$$ $$25.5 \space g \space NH_3 \times \frac{1 \space mole \space NH_3 }{ 17.03 \space g \space NH_3 } \times \frac{ 5 \space moles \space F_2 }{ 2 \space moles \space NH_3 } \times \frac{ 38.00 \space g \space F_2 }{1 \space mole \space F_2 } = 142 \space g \space F_2$$ d. $NH_3$ : 17.03 g/mol $$\frac{1 \space mole \space NH_3 }{ 17.03 \space g \space NH_3 } \space and \space \frac{ 17.03 \space g \space NH_3 }{1 \space mole \space NH_3 }$$ $N_2F_4$ : ( 14.01 $\times$ 2 )+ ( 19.00 $\times$ 4 )= 104.02 g/mol $$\frac{1 \space mole \space N_2F_4 }{ 104.02 \space g \space N_2F_4 } \space and \space \frac{ 104.02 \space g \space N_2F_4 }{1 \space mole \space N_2F_4 }$$ $$3.40 \space g \space NH_3 \times \frac{1 \space mole \space NH_3 }{ 17.03 \space g \space NH_3 } \times \frac{ 1 \space mole \space N_2F_4 }{ 2 \space moles \space NH_3 } \times \frac{ 104.02 \space g \space N_2F_4 }{1 \space mole \space N_2F_4 } = 10.4 \space g \space N_2F_4$$