Question

Can someone please help????
If 25.0 grams of the base were used in the neutralization reaction in question 7, ow many grams of acid must have been used to neutralize the base? Btw. answer to #7 was the acid: HF; the base: Al(OH)3. this forms AIF3

Answer 1

First figure out how many moles of base you have. You can see that there are 3 equivalents of base produced by Al(OH)3 (that is, 3 OH- ions), but only 1 equivalent of acid produced by HF (that is, 1 H+ ion). This means that for however many moles of AL(OH)3 you have, you are able to neutralize 3 times as many moles of acid (if you write out and balance the neutralization reaction you'll see this is the case). Once you figure out how many moles of acid there are, multiply by the molar mass to find how many grams were used!

Answer 2

Why don't you write down the equation? How do we know what we're dealing with here.

Answer 3

3HF + Al(OH)3 ----> 3H2O + AlF3
there's the equation. i still don't understand how to do it though

Answer 4

So first you need to identify which is the base you're starting with.
And then identify the acid that you're forming. Can you tell me which one is the base and which is the acid that you're forming

\(\sf \color{blue}{Reactant~ base}\):
\(\sf \color{Red}{Product~acid}:\)

Answer 5

Next, you need to use the given information.
For these types of problems, what you need to do is ALWAYS write down the balanced chemical equatin. FIRST STEP! Always!!
So, you have: \(\sf \color{green}{3HF + ~Al(OH)_3 \rightarrow 3~H_2O + AlF_3}\)

Answer 6

Next, write down what you're starting with.

and write down a plan of what you have and what you want: \(\sf \color{red}{have}~\rightarrow ~ \color{blue}{want}\)

Next, write the full steps of the plan HOW to get there:

have: \(\sf \color{blue}{25.0~g~of~base} \rightarrow mol~of~base \rightarrow \color{red}{mole~want} \rightarrow ~ \color{red}{grams~of~acid}\)

Answer 7

First step is to take the 25 g of base and divide it by the MASS of the base (molar mass).

Remember that molar mass is \(\sf \large \frac{grams}{mol}\)
But you want to get rid of GRAMS to get into moles. So you're dividing:

\(\sf \color{blue}{25~\cancel{grams~base}} \times \large \frac{moles~of~base}{\cancel{grams~of~base}}\)

Notice how they both cancel out? And you're left with only moles? Then you can convert using this same technique, to get your GRAMS of acid.

Answer 8

THANK YOU