Question

How many grams of Mg are needed to react completely with 3.80 L of a 2.50 M HF solution?
Mg + 2HF yields MgF2 + H2

Answer 1

@Photon336 can you help me with this last one please?

Answer 2

this is tricky but the information we need is hidden in the information

Answer 3

\[Molarity_{HF}*(Volume_{HF}) = Moles~HF\]

Answer 4

the molarity is the number of moles per liter. if we know the volume we can easily get the number of moles of HF by multiplying the following.

\[\frac{ Moles }{ L }*L = Moles_{HF}\]

So try this and tell me what you get

Answer 5

well the L would cancel each other out

Answer 6

exactly

Answer 7

and we would be left with the number of moles of HF but we are not done quite yet

Answer 8

try to see if you can fin the number of moles of HF @sammyalabamy

Answer 9

it would be 2.5

Answer 10

\[2.50(\frac{ Moles }{ L })*3.80L = 9.5~moles \]

Answer 11

Thank you so much @Photon336

Answer 12

that's just the first part @sammyalabamy now we need to find the number of moles of Mg. so now. my question is the reaction balanced and how do you know?

Answer 13

there are 2 of H and F on each side as well as only 1 Mg

Answer 14

the reaction is balanced because the same number of moles of each atom are on both sides, this is important because we can only do stoichiometry with balanced equations.

now let's look at the molar ratio of HF to Mg we know that

we need 2Moles of HF to react with 1 mole of Mg right?

Answer 15

yes

Answer 16

so now to find out how many moles of magnesium we need we multiply the number of moles by the molar ratio of the two.


9.5moles HF *( 1 mole Mg/ 2 moles HF = 4.8 moles of Mg needed

Answer 17

thanks