1. Calculate the weight of KCLO3 that would be required to produce29.52L of oxyg

Question

1. Calculate the weight of KCLO3 that would be required to produce29.52L of oxygen measured at 127 degC and 1atm.2KCLO3(s)....heat>2KCL(s)+3O2(g) ... Answer is 73.5 grams,please show me how you got there.

2. What volume of dry gaseous HBr at STP can be obtained from thereaction of 81.2g of PBr3 with excess water?PBr3(l)+3H2O(l)...>H3PO3(l)+3HBr(g) ... Answer is 20.2L, pleaseshow me how you got there.

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Explanation / Answer

1 - 2KCLO3(s)....heat>2KCL(s)+3O2(g) 29.52L of oxygen measured at 127 degC and 1atm. Let's find how much oxygen will be produced in mol, using the gasesequation: P = pressure V = volume n = number of mol R = constant T = temperature P . V = n . R . T P = 1 atm V = 29,52 L n = ? R = 0,082 atm . L / (mol . k) T = 127oC = 400 K (TKelvin =TCelcius + 273) So.. P . V = n . R . T (1) . (29,52) = n . (0,082) . (400) n = 0,9 mol So.. The reaction is: 1 - 2KClO3(s)....heat>2KCl(s)+3O2(g) So 2 mol of KClO3 produces 3 mol of O2. Molar mass of KClO3 is = 122,5 g/mol(approximately). So if 1 mol of KClO3 has 122,5 g, 2 mol have [2 x(122,5)] = 245g. So if 2 mol of KClO3 produces 3 mol of O2,that means 245 g of KClO3 produces 3 mol ofO2. But we calculated that just 0,9 mol of O2 will beproduced. Then.. 245 g of KClO3 produces 3 mol of O2. x g of KClO3 produces 0,9 mol of O2. Just make a rule of three: 245 g of KClO3 - 3 mol of O2x g ofKClO3 - 0,9 mol of O2 x = 73,5g of KClO3 2. What volume of dry gaseous HBr at STP can beobtained from the reaction of 81.2g of PBr3 with excesswater?PBr3(l)+3H2O(l)...>H3PO3(l)+3HBr(g)... Answer is 20.2L, please show me how you got there. If the water is in excess, it doesn't matter. We just have toconsider that all PBr3 will react (but no all thewater). So... 1 mol of PBr3 produces 3 mol of HBr. Molar mass of PBr3 is 270,7 g/mol. So 270,7 g of PBr3 produces 3 mol of HBr. In STP, 1 mol of any gas has 22,4 L. So for 3 mol we would have: (3x 22,4) = 67,2 L. So 270,7 g of PBr3 produces 67,2 L of HBr. If we react just 81,2 g of PBr3 , we would produce x Lof HBr. Rule of three. 270,7 g of PBr3 - 67,2 L of HBr 81,2 g of PBr3- x L of HBr x = 20,15 L (or 20,2approximately)

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