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    Boron trichloride (BCl3) can be prepared as shown by the following equation.B2O3(s) + 3C(s) + 3Cl2(g) 2BCl3(g) + 3CO(g) A sample of boron oxide (B2O3) was reacted completely with carbon and chlorine. The two gases produced occupied a total volume of 5000 cm3 at a pressure of 100 kPaand a temperature of 298 K. Calculate the mass of boron oxide that reacted. Give your answer to 3 significant figures. (The gas constant R = 8.31 J K–1 mol–1)I have tried to work this out and I looked on the mark scheme which is :P = 100 000 (Pa) and V = 5.00 x 10-3 (m3)n = PV/RT = 100 000 x 5.00 x 10-3/8.31 x 298 = 0.202 moles (of gas produced)Therefore 0.202/5 = 0.0404 moles B2O3 Mass of B2O3 = 0.0404 x 69.6= 2.81 (g)I understand all of this except i don't understand why you divide 0.202 by 5. Why do i divide by 5?

    In this reaction, 0.202 moles of gas are produced. If you look at the equation 2BCl3 and 3CO are produced which in total give 0.202 moles. The ratio of these gases to the moles of B2O3 produced is 5:1. (2 of BCl3 and 3 of CO) Therefore the moles of B2O3 must be 0.202/5 mol.
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