Chemistry For Engineering-1421-105 Q1 :How many grams of sodium phouphate are re

Question

Chemistry For Engineering-1421-105 Q1 :How many grams of sodium phouphate are required to prepare 85o .os m solution whose concentration is 7.461 M 02- Octane i s a component of gasoline. Complete combustion of octane yields H,O. Incomplete combustion produces Co and HO. In a certain run. uned in an engine. The total mass of Co, CO, and 1H-0 produced is 14.053 kg. Calculate the efficiency of the process, that is, calculate the 2.345 gal of octane is b fraction of octane converted to CO,. The density of octane is 2.65 kg/gal Q3- What are the definition of actual yield, solubility, and gravimetric analysir Q4- Calculate the mass of aluminum chloride produced by the reaction betw aluminum hydroxide and magnesium chloride, use the mass of alumi hydroxide 82.098g? Q5-A certain metal M forms a 60.1 percent of chloride by mass. What chemical formula of the compound? Ifuric acid are needed to reduc of a 0.345 M barium hydroxide?

Explanation / Answer

Q1. grams of sodium phosphate Na3PO4

molarity = grams/molar mass x volume

with,

molarity = 7.461 M

Volume = 850.05 ml = 0.85005 L

molar mass = 163.94 g/mol

we get,

grams of Na3PO4 needed = 7.461 M x 0.85005 L x 163.94 g/mol

                                           = 1039.744 g

Q2. mass of octane taken = 2.345 gallon x 2.65 kg/gallon = 6.214 kg

mass of products = 14.053 kg

Theoretical yield of CO2 = 6.214 kg x 16 x 44/114.23 g/mol x 2 = 19.15 kg

efficiency of process = 14.053 kg x 100/19.15 kg = 73.40%

Q3. actual yield = mass of product obtained from running actual experiment

solubility = the amount of solute dissolved in a solvent and it remains in a solution form without getting precipitated out from solution.

gravimetric analysis = quantitative identification in terms of weight of the compound in a sample.

Q4. question not displayed fully

Q5. mass% Cl in compound = 60.1%

molecular formula possible = CaCl2

Test-3

Q1. heat Q = mCpdT

m = 23 g

Cp = 0.72 J/g.oC

dT = 78 - 22 = 56 oC

we get,

heat Q = 23 x 0.72 x 56 = 927.36 J

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