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2 years ago

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]2 years 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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c3H8 + 5O2 = 3CO2 + 4H2O When 44.0 grams of propane (C3H8) under goes complete combustion, how many grams of water will be produ

Umnica [9.8K]

<u>Answer:</u> 72 grams of water will be produced.

<u>Explanation:</u>

To calculate the number of moles, we use the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

Mass of propane = 44 grams

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Putting values in above equation, we get:

$\text{moles of propane}=\frac{44g}{44g/mol}=1mole$

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By Stoichiometry of the reaction,

1 mole of propane produces 4 moles of water.

So, 1 mole of propane will produce = $\frac{1}{1}\times 4=4moles$ of water.

Now, to calculate the amount of water, we use equation 1, we get:

Molar mass of water = 18 g/mol

$4mol=\frac{\text{Mass of water}}{18g/mol}$

Mass of water produced = 72 grams

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