baf₂ → f₂ + ba\nhow many grams of fluorine, f₂, form from 12.0 moles of barium fluoride, baf₂?\nstep 1: how…

baf₂ → f₂ + ba\nhow many grams of fluorine, f₂, form from 12.0 moles of barium fluoride, baf₂?\nstep 1: how many moles of fluorine form?\n12.0 mol baf₂ → ? mol f₂

baf₂ → f₂ + ba\nhow many grams of fluorine, f₂, form from 12.0 moles of barium fluoride, baf₂?\nstep 1: how many moles of fluorine form?\n12.0 mol baf₂ → ? mol f₂

Answer

Explanation:

Step1: Determine mole - ratio

From the balanced chemical equation $BaF_2\rightarrow F_2 + Ba$, the mole - ratio of $BaF_2$ to $F_2$ is 1:1. So, if we have 12.0 mol of $BaF_2$, the number of moles of $F_2$ formed is also 12.0 mol because $n_{F_2}=n_{BaF_2}\times\frac{1\ mol\ F_2}{1\ mol\ BaF_2}=12.0\ mol\times1 = 12.0\ mol$.

Step2: Calculate mass of $F_2$

The molar mass of $F_2$ is $M = 2\times19.0\ g/mol=38.0\ g/mol$. Using the formula $m = n\times M$, where $n = 12.0\ mol$ and $M = 38.0\ g/mol$, we get $m=12.0\ mol\times38.0\ g/mol = 456\ g$.

Answer:

456 g