RDV: Amount of Substance

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

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.

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.

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.

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?

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?

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?

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.

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.

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.

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.