4.48 mol Cl2. A reaction that produces 0.35 kg of BCl3 will use 4.48 mol of Cl2.
(a) The <em>balanced chemical equation </em>is
2B + 3Cl2 → 2BCl3
(b) Convert kilograms of BCl3 to moles of BCl3
MM: B = 10.81; Cl = 35.45; BCl3 = 117.16
Moles of BCl3 = 350 g BCl3 x (1 mol BCl3/117.16 g BCl3) = 2.987 mol BCl3
(c) Use the <em>molar ratio</em> of Cl2:BCl3 to calculate the moles of Cl2.
Moles of Cl2 = 2.987 mol BCl3 x (3 mol Cl2/2 mol BCl3) = 4.48 mol Cl2
Due to the law of conservation of mass, _g K + 16g O=94g KO.
So, 94-16=78
78 is your answer
24.305 g +28.08 g + 3(15.999 g)=100.382
219 g MgSIO3 (1 mole MGSIO3/100.382g)=2.181 mol MgSIO3
the answer is 2.18 mol MgSIO3