RDV: Amount of Substance

How many particles make up 0.67 moles of Ammonia (NH3)?

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Answer explanation

To find the number of particles in 0.67 moles of NH3, use Avogadro's number (6.022 x 10^23 particles/mole). Multiply: 0.67 moles x 6.022 x 10^23 particles/mole = 4.03 x 10^23 particles.

Calculate the number of moles of 23g of Gold. Give your answer to 2 significant figures.

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Answer explanation

To find moles, use the formula: moles = mass (g) / molar mass (g/mol). Gold's molar mass is 197 g/mol. Thus, moles = 23 g / 197 g/mol = 0.116 moles, rounded to 0.12 moles (2 significant figures).

Calculate the number of moles of 200cm3 of 0.35moldm-3 HCl.

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Answer explanation

To find the number of moles, use the formula: moles = concentration (mol/dm³) × volume (dm³). Convert 200 cm³ to dm³ (0.2 dm³). Thus, moles = 0.35 mol/dm³ × 0.2 dm³ = 0.07 moles.

Convert 0.5mol/dm3 of sodium hydroxide into g/dm3.

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Answer explanation

To convert 0.5 mol/dm³ of sodium hydroxide (NaOH) to g/dm³, use the molar mass of NaOH (40 g/mol). Multiply: 0.5 mol/dm³ × 40 g/mol = 20 g/dm³. Thus, the answer is 20 g/dm³.

What volume of H2 is produced when 12g of Potassium reacts with Water at 100kPa of pressure and 298K?

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Answer explanation

Potassium reacts with water to produce H2 gas. The molar mass of K is 39.1 g/mol, so 12 g yields 0.31 mol. At 298K and 100kPa, using the ideal gas law, the volume of H2 produced is approximately 7.6 L.

A compound contains 23.3% Magnesium, 30.7% Sulfur and 46.0% Oxygen. What is the empirical formula of this compound?

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Answer explanation

Convert percentages to grams (assuming 100g total): 23.3g Mg, 30.7g S, 46.0g O. Convert to moles: Mg=0.958, S=0.960, O=2.875. Divide by the smallest number of moles (0.958) to get ratios: Mg=1, S=1, O=3. Empirical formula is MgSO3.

A hydrocarbon combusts completely to make 0.845g of CO2 and 0.173g of H2O. What is the empirical formula of the hydrocarbon?

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Answer explanation

To find the empirical formula, convert grams of CO2 and H2O to moles. CO2: 0.845g/44.01g/mol = 0.0192 mol C; H2O: 0.173g/18.02g/mol = 0.0096 mol H. The ratio C:H = 0.0192:0.0096 = 2:1. Empirical formula is C2H5.

In a reaction involving the complete combustion of Calcium 32.6g of Calcium Oxide was produced. The theoretical mass is 47.6g. Calculate the percentage yield of this reaction.

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Answer explanation

Percentage yield = (actual yield / theoretical yield) x 100. Here, (32.6g / 47.6g) x 100 = 68.4%. Thus, the percentage yield of the reaction is 68.4%.

Calculate the atom economy in the extraction of Iron from the reaction Fe2O3 + 1.5C → 2Fe + 1.5CO2.

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Answer explanation

Atom economy = (mass of desired product / total mass of reactants) x 100. Here, desired product is 2Fe (mass = 112g) and total reactants = 160g (Fe2O3 + 1.5C). Atom economy = (112/160) x 100 = 70%. Thus, atom economy is 70%.

Calculate the concentration of KOH if 18.3cm3 of 0.25moldm-3 HCl was required to neutralise 25cm3 of KOH.

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Answer explanation

To find KOH concentration, use the neutralization equation: HCl + KOH -> KCl + H2O. Moles of HCl = 0.25 mol/dm³ * 0.0183 dm³ = 0.004575 mol. Since the ratio is 1:1, moles of KOH = 0.004575 mol. Concentration = moles/volume = 0.004575 mol / 0.025 dm³ = 0.183 mol/dm³.