Question

How many grams of Mg are needed to react completely with 2.5 L of a 3.8 M HF solution? Mg + 2 HF --> MgF2 + H2

Answer 1

@Compassionate @Luigi0210 @e.mccormick

Answer 2

@Abhisar can u explain this too me?

Answer 3

@shamim

Answer 4

From the given equation,

\(\large \tt \hspace{75pt}Mg + 2 HF \Rightarrow MgF_2 + H2\)

We can say that 1 mole of Mg i.e 24.3 g of Mg is needed to completely react with 2 moles of HF. Thus by unitary method it is obvious that 12.15 g of Mg is needed to react with 1 mole of HF.

Next, we will calculate the number of moles of HF present in 2.5 L of 3.8 M of HF solution. This can be done by using the equation,

\(\tt \hspace{85pt}\boxed{ Conc=\large \frac{Moles}{Volume}}\)

Plugging in the values we get,

\(\tt 3.8 = \Large \frac{Moles}{2.5}\\ \Rightarrow Moles=3.8\times 2.5=9.5 \)

Finally, as we know that 12.15 g of Mg is required to react completely with 1 mole of HF so \(\tt 12.15 \times 9.5 =~?\) g of Mg will be required to compleltely react with 9.5 moles of HF or 2.5 L of 3.8 M HF.