is anyone here a chemistry genius, im taking a test and I'm about to flunk the whole class...

I am good as long as it is not practical.... Am good with theories tho.... And also in IUPAC....

the question is "if 0.940 mole of HBr in solution is added to 16.5 grams of MgO in a beaker and then is heated until dry. what are the expected contents of the beaker" idk anything about chemistry, my teacher doesnt actually teach. if you cant answer it is fine, i think ill do a bit more research on it, thank you anyways

Very tough since it's a reaction between acid and base... I'll try tomorrow by opening my notes and also try asking my frndz... Since it's midnight.....

That's an acid-base reaction I'm assuming? So between those elements you first need to balance the chemical equation.
Which is:
MgO + 2HBr --> MgBr2 (s) + H2O (l)

Since it's heated until dry that means they're only asking you for the amount of solid made from the reaction.
Since you have 0.94 mole of HBR, according to stoichiometry it takes 2 moles of HBr to make 1 equivalent of MgBr2

Then you just use that stoichiometric ratio to convert the amount of moles you have to products

0.940 mol HBr x (1 mol MgBr2/ 2 mol Hbr) = 0.47 mole MgBr2

And if they want the answer to be in mass, just convert the moles to it's molecular weight

0.47 mol MgBr2 x (184.1g MgBr2 / 1 mole MgBr2) = 86.5g MgBr2, assuming 100% conversion to products

Let me know if you're working on different material than unit conversions though in case the question is asking for something else.

I’m smart asf when it comes to science because I cheat using google

Oh you know what ignore me, I didnt read your question entirely and ignored that HBr is a limiting reacted, give me one second

Ok so once again starting off with your balanced equation of

MgO + 2HBr --> MgBr2 (s) + H2O (l)

What the problem wants you to determine first is what is the limiting reactant.

With 0.94mol HBr2 convert it to grams to see which reagent is in excess or you can convert 16.5 g of MgO to moles.

I'll start with 0.94 mol HBr x (80.91g HBr/ 1 mol HBr) = 76.1g HBr
Or 16.5g MgO x (1 mol MgO /40.3g MgO) =0.41 mol MgO

So if you have 0.94 mol Hbr and only 0.41 mol MgO, according to the stoichiometric ratio you need 1 MgO mole to react with 2 Mole Hbr.

Determine which is the limiting reactant:

0.94 mol HBr2 x (1 mol MgO/ 2 mol Hbr) = 0.47 mol MgO, which you do not have since 16.5g MgO is only around 0.41 mol
Thus MgO is your limiting reactant

Now convert the amount of product yielded with the limiting reactant by the same way of using stoichiometric ratios and convert to grams if they want mass.

Thank youuuuu, i appreciate ittt<3333

i tried to cheat but my teacher made her own test :( i've never been good at chemistry stuff, its always so complicated, like why is there math in it?

i seriously love you, thank youu, i hope you have the best day ever!

OMG FRR I HATE WHEN SUM MY TEACHERS DO THAT SHIT OOH
spain with out the s