3. Consider the following balanced equation. If you were to combine 25.0 grams o

Question

3. Consider the following balanced equation. If you were to combine 25.0 grams of hydrogen gas and 25.0 grams of oxygen gas in a reaction vessel, what is the limiting reactant? Show your work. 2H2 + O2 2H2O 4. Consider the following balanced equation. Calculate the theoretical yield of calcium chloride that can be produced from 50.0 mL of 2.5 M HCI and 5.50 grams of calcium carbonate. 2 HCl + CaCO3 CaCl2 + H2O + CO2 5. How many grams of (NH4)2S04 are in 250 mL of a 0.15 M solution? 6. A solution of NaOH is made by dissolving 25.0 grams into 450.0 mL. The density of water is 0.99427 when it is at the temperature of the lab. Calculate the molality of the NaOH solution. 7. In lab, 10.0 mL of a 2.5 M stock solution of HCI is diluted to 50.0 mL using a volumetric flask. 5.0 mL of the resulting solution is then diluted to 20.0 mL. What is the concentration of the final solution?

Explanation / Answer

5.

Moles of (NH4)2SO4 = molarity * Volume / 1000

n = 0.15 * 250 / 1000

n = 0.0375 mol

And molar mass of (NH4)2SO4 = 2 (14) + 8(1) + 1 (32) + 4 (16) = 132 g/mol

Formula,

Number of moles = mass / molar mass

0.0375 = Mass / 132

Therefore, Mass of (NH4)2SO4 = 4.95 g.

6.

Mass of NaOH = 25.0 g.

Molar mass of NaOH = 40.0 g/mol

Moles of NaOH = mass / molar mass = 25.0 / 40.0 = 0.625 mol

Mass of solvent = density * volume = 0.99427 * 450.0 = 447.4 g. = 0.4474 kg.

Therefore,

Molality = moles of solute / mass of solvent in kg

m = 0.625 / 0.4474

m = 1.40 m

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