We are provided with
4.35 g of phosphoric acid
5.25 g of KOH
3.15 g of K3PO4 produced
The reaction formula is
H3PO4 + 3KOH => K3PO4 + 3H2O
Initially, convert the given masses into moles.
Then, identify the limiting reactant. Afterward, calculate the maximum K3PO4 that can be generated from the limiting reactant.
Finally, compute the percent yield by dividing the actual yield by the theoretical yield.
Responses: a. 1.28 mol/L; b. 17.0 %; c. 0.0227; d. 1.29 mol/kg Explanation: a. Molar concentration: c = moles/litres. Moles = 167 × 1/159.61. After performing the calculation, Moles = 1.046 mol. Litres = 820 × 1/1000. Hence, Litres = 0.8200 L. Calculating the molar concentration gives c = 1.046/0.8200, resulting in c = 1.28 mol·L⁻¹. b. Percent by mass: Mass % = mass of solute / mass of solution × 100 %. Mass of solution = volume × density, therefore, Mass of solution = 820 × 1.195. By calculating this, Mass of solution = 979.9 g. Thus, Mass % = 167/979.9 × 100, which results in Mass % = 17.0 %. c. Mole fraction: χ = moles of solute / (moles of solvent + moles of solute). Mass of solvent = mass of solution – mass of solute; namely, Mass of solvent = 979.9 – 167. Converting this to moles gives Moles of water = 812.9 × 1/18.02, which results in Moles of water = 45.11 mol. The total moles are 1.046 + 45.11, leading to Total moles = 46.16 mol. Finally, the mole fraction is calculated as χ = 1.046/46.16, equating to χ = 0.0227. d. Molal concentration: b = moles of solute / kilograms of solvent. Mass of solvent = 812.9 g = 0.8129 kg. Therefore, the molal concentration yields: b = 1.046/0.8129 = 1.29 mol/kg.
The mixture’s density is 1.57 g/cm³.
Step 1: Determine the mass of the butter.
Step 2: Determine the mass of the sand.
Step 3: Determine the density of the mixture.
Total mass = 0.860 g + 2.28 g = 3.14 g.
Total volume = 1 cm³ + 1 cm³ = 2 cm³
M1V1 = M2V2
(2.50)(100.0) = (0.550)V2
V2 = 455mL
From 100.0 mL of 2.50 M KBr, you can prepare 455 mL of 0.550 M solution.
