Answer
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.
Work Step by Step
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
$$ \frac{1 \space mole \space N_2F_4 }{ 104.02 \space g \space N_2F_4 } \space and \space \frac{ 104.02 \space g \space N_2F_4 }{1 \space mole \space N_2F_4 }$$
$$ 3.40 \space g \space NH_3 \times \frac{1 \space mole \space NH_3 }{ 17.03 \space g \space NH_3 } \times \frac{ 1 \space mole \space N_2F_4 }{ 2 \space moles \space NH_3 } \times \frac{ 104.02 \space g \space N_2F_4 }{1 \space mole \space N_2F_4 } = 10.4 \space g \space N_2F_4 $$
