11) How many moles of Na\' are in 2 g of Naz504? (1) 12) How many moles of Naz50

Question

11) How many moles of Na' are in 2 g of Naz504? (1) 12) How many moles of Naz504 is needed to produce 10 g of Na product? (2) 13) How many grams of Ba product is formed if 2.5 g of BaCl2 is used in the reaction? 14) What is the theoretical yield of Ba product, if 250 g of Na2S04 reacts with 250 g of BaCl2? Which of the two reactants is the limiting reagent? (2) 15) What is the percentage yield if only 95 g of Na product was recovered after consuming all 142 g of NazS04? (2) 16) Show Oxidation Numbers of each element in the reactant and product- (2) 17) Tell whether this reaction is Neutralization, Precipitation, Redox, Non Redox reaction. (1/2) snonse in this aniz reflects mv true understanding 2

Explanation / Answer

11. Number of moles of Na2SO4 = mass/molar mass = 2/142.04 = 0.0014 mole

Number of moles of Na = 2 × number of moles of Na2SO4 = 2 × 0.0014 = 0.0028 moles

12. Number of moles of Na in 10 g = 10/23 = 0.4348

2 moles of Na are produced by 1mole of Na2SO4

So, 0.4348 would have been produced from 0.4348/2 = 0.2174

13. Since, 208.23 g BaCl2 produces 137.33 g

So, 2.5 BaCl2 ............. 137.33 × 2.5/208.23 = 1.65 g

14. Number of moles of Na2SO4 = 250/142.04 = 1.76

Number of moles of BaCl2 = 250/208.23 = 1.20

Limiting agent: BaCl2

Number moles of BaSO4 = number moles of BaCl2 = 1.2

Mass of BaSO4 = number moles × molar mass =1.2 × 233.38 = 280.056 g

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