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tamaranim1 [39]
1 year ago
7

How many mL of a 0.193 M aqueous solution of ammonium bromide, , must be taken to obtain 8.42 grams of the salt?

Chemistry
1 answer:
PilotLPTM [1.2K]1 year ago
8 0

Answer:

445 mL NH₄Br

Explanation:

(Step 1)

Convert grams ammonium bromide (NH₄Br) to moles using its molar mass.

Molar Mass (NH₄Br): 14.007 g/mol + 4(1.008 g/mol) + 79.904 g/mol

Molar Mass (NH₄Br): 97.943 g/mol

8.42 grams NH₄Br                1 mole
-----------------------------  x  ------------------------  =  0.0860 moles NH₄Br
                                         97.943 grams

(Step 2)

Calculate the volume of NH₄Br using the molarity equation. Then, convert liters (L) to milliliters (mL).

Molarity (M) = moles / volume (L)

0.193 M = 0.0860 moles / volume

(0.193 M) x volume = 0.0860 moles

volume = 0.445 L

1,000 mL = 1 L

0.445 L NH₄Br           1,000 mL
------------------------  x  ------------------  =  445 mL NH₄Br
                                        1 L

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I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
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Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

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