Question

2. What is the percent yield or if the reaction of 25.0 g P. and 1596. produces 104 g PCI? P.(s) + Cl2(o) PCA) If the reaction produces 5.3 mg of carbon dioxide how many grams of water are produced? d. How many grams of oxygen are needed to react with 6.00 x 10molecules of CH? 3. How many grams of aluminum chloride are formed when 30.0 grams of aluminum are reacted with gas? a. Write the balance equation b. Calculate the number of grams of aluminum chloride produced 4. Titanium(IV) oxide (TIO,) is a white substance produced by the action of sulfuric acid on the ilmenite (FeTiO3): FeTio + H SO. - TIO, + Fesos + H2O a. How many kg of titanium(IV) oxide are produced from 8.00 X 102 kg of Fetios? b. How many kg of FeSO. are produced from 8.00 X 102 kg of FeTIO,? C. How many kg of Fetio, are required to produce 2.00 kg of TiO2?

Answer 1

The image is very unclear to understand the letters and numbers. Anyhow I managed to read the first one (number 2) and solved it.

2.c. From where CO2 and water comes?

2.d. What is the hydrocarbon compound in the question. It is not clear at all.

Next time please upload a clear image.

The balance chemical reaction of Pu and Clz are shown below Putolla = 4P4 3. The reason the shows that the one more of Py reacts us 6moles of Clz gives 4 moles of pulz. Now the molar mass of Pu= 123.88 g/mol cz has molar mass = 70.906 g/mol Pelz has molar mass = 137.33g /mol. 250g of Pu reacts with glisg chin gives 104108 of pus. We need to find out the percent of yield of the reaction- For that it is necessary to find öntite number of moles offt of the reactants and products Number of mous Can be calculated by using following formerta. ... weight of the substance Number of moles of substance = mótar man of the substance : 2500g Pu = 250 g =0:2018 mol. . 123.888/mol 91.5g Cl2 = als het mot = 1,2904 mot 10o = 0.7573 According to the balanced chemical equation 0 2018 mol Pu will react with (6X0-2018) mol=112108 mo) of chlorine to give (4x0.2018) mol= 0: 8432 mol of PC3. But in the reaction 07573 mol of PU3 (10410g) is formed Therefore experimental yield> 0.7573 mol and theoretical yield is 0.8432 mol. % of yield of the reaction = molog experimental yiedd edd x 100% . mol of theoretical yield 1 -0.7573 mol x 100% = 0.8432 mol To 40 g Pu3 = 104 = 0.7573 89.8%