# Quiz 3 - Chemical Calculations | Chemical Calculations

1

How many moles of ions are in 75.0 mL of a 1.20 M aqueous solution of sodium sulfate?

• Sodium sulfate (Na2SO4) dissociates into 2 Naand 1 SO42-, giving 3 ions per formula unit.
• Moles of Na2SO4 = 1.20 M x 0.075 L = 0.090 mol.
• Total moles of ions = 0.090 mol x 3 = 0.270 mol.
2

When 3.00 g of NH3 (M = 17.03) reacts with 18.0 g of HCl (M = 36.46), what is the theoretical yield of NH4Cl?

• Moles of NH3 = $\frac{3.00}{17.03}$ = 0.176 mol, moles of HCl = $\frac{18.0}{36.46}$ ​= 0.493 mol. NH3 is the limiting reactant.
• Theoretical yield of NH4Cl = 0.176 mol x 53.49 g/mol = 9.42 g.
3

200.0 mL of aqueous nitric acid is mixed with 300.0 mL of 0.200 M Mg(NO3)2 solution to give a solution with a nitrate ion concentration of 0.500 M. What is the concentration of the nitric acid?

• Total volume = 500.0 mL = 0.500 L
• Moles of NO3- from Mg(NO3)2 = 0.300 L x 0.200 M x 2= 0.120 mol.
• Moles of NO3- required = 0.500 M x 0.500 L = 0.250 mol.
• Moles of NO3- from HNO3 = 0.250 mol - 0.120 mol = 0.130 mol ⇒ Concentration of HNO3 =  = 0.650M.
4

What volume of 18.0 M sulfuric acid must be diluted to 250.0 mL to afford a 0.55 M solution of sulfuric acid?

• Dilution equation C1V1 = C2V2​, where C1 = 18.0 M, C2 = 0.55 M, V2 = 250.0 mL
• V1 = $\frac{{\mathrm{C}}_{2}{\mathrm{V}}_{2}}{{\mathrm{C}}_{1}}$ =  = 7.64 mL
5

Iron(III) sulfide reacts with gaseous hydrogen chloride to produce iron(III) chloride and hydrogen sulfide. How much iron(III) sulfide would be required to react with 1.5 mol HCl?

• The balanced reaction is Fe2S3 + 6 HCl → 2 FeCl+ 3 H2S.
• Moles of  Fe2S3 =  = $\frac{1.5}{6}$ = 0.25 mol.
6

How much Sr(OH)2.8H2O (M = 265.76) is needed to prepare 250.0 mL of solution in which [OH] = 0.100 M?

• First, calculate the moles of OH- ions required: moles of OH- = M x V = 0.100 M x 0.250 L = 0.025 mol
• Since Sr(OH)2 dissociates completely in water to produce 2 OH- ions per formula unit:
moles of Sr(OH)2 =  =  = 0.0125 mol
• Now, calculate the mass of Sr(OH)2.8H2O needed: mass = 0.0125 mol x 265.76 g/mol = 3.32 g
7

What is the concentration of calcium ions in 350. mL of an aqueous solution containing 7.50 g CaCl2?

• Molar mass of CaCl2 =  110.98g/mol
• Calculate moles of CaCl2:  = 0.0676 mol
• Calculate concentration of Ca2+$\frac{{\mathrm{n}}_{{\mathrm{CaCl}}_{2}}}{\mathrm{V}}$ =  = 0.193 M
8

0.422 g of an element Z reacts with oxygen to form 0.797 g of the oxide Z2O3. What is the element Z?

• Calculate the mass of oxygen in the oxide: Mass of O = Mass of Z2O3 − Mass of Z = 0.797 g − 0.422 g = 0.375 g
• Calculate the moles of oxygen: moles of O =  = 0.02344 mol
• Since the formula of the oxide is Z2O3, the moles of Z should be: moles of Z = $\frac{2}{3}$ x moles of O = 0.01563 mol
• Calculate the molar mass of Z: MZ =  =  = 27.00 g.mol-1

The element with a molar mass of approximately 27.00 g/mol is aluminum (Al).

9

In a sample consisting of 1.00 mol NaBr and 0.300 mol KI, what is the mass percent of iodine?

• Calculate the mass of I: Mass of KI = 0.300 mol × 166.00 g/mol = 49.80 g; Mass of I in KI = 0.300 mol × 126.90 g/mol = 38.07 g
• Calculate the total mass of the sample: Mass of NaBr = 1.00 mol × 102.89 g/mol = 102.89 g; Total mass = 102.89 g + 49.80 g = 152.69 g

• Calculate the mass percent of iodine: Mass percent of I = (38.07 g / 152.69 g) × 100 ≈ 24.9%

10

What is the concentration of chloride ions in a solution formed by mixing 150. mL of a 1.50 M NaCl solution with 250. mL of a 0.750 M MgCl2 solution?

• Calculate moles of Cl⁻ from NaCl:
Moles of NaCl = 1.50 M × 0.150 L = 0.225 mol. Each mole of NaCl provides 1 mole of Cl⁻ ⇒ Moles of Cl⁻ from NaCl = 0.225 mol
• Calculate moles of Cl⁻ from MgCl2:
Moles of MgCl2 = 0.750 M × 0.250 L = 0.1875 mol. Each mole of MgCl2 provides 2 moles of Cl⁻ ⇒ Moles of Cl⁻ from MgCl2 = 0.1875 mol × 2 = 0.375 mol
• Total moles of Cl⁻ = 0.225 mol + 0.375 mol = 0.600 mol
• Total volume = 150 mL + 250 mL = 400 mL = 0.400 L
• Calculate the concentration of Cl-: Concentration of Cl- =   =  = 1.50 M