chloride, AlCl3. What volume of hydrogen at STP will be produced when 35.0 grams of aluminum reacts with HCl?
it rather is an oxidation help reaction it must be belanced has one so first you should calculate the oxidation numbers (o.n.) of the species: eq: Al(s) + HCl ? H2 (g) + AlCl3 o.n.: 0......+a million -a million ......0......... +3 -a million The oxidation style of H modifications of +a million to 0 so its a help reaction. The oxidation style of Al modifications of 0 to +3 so its a oxidation reaction. calculate the balanced equation (Al(s) ? AlCl3 + 3e-) x 2 (2e- + 2HCl ? H2) x 3 --------------------------------------..... 2Al(s) + 6HCl ? + 2AlCl3 + 3 H2
First, write the balanced equation: 2Al + 6HCl -- 2AlCl3 + 3H2. Then look up the atomic weight of aluminum. It's 26.98. (Since the given amount of aluminum has three significant figures, it's best to do the calculations with four or five and then discard the extra one or two in the answer.) So 35.0g of aluminum is 35.0/26.98 = 1.297 mole of aluminum. From the reaction, each mole of aluminum yields 3/2 mole of hydrogen. So the amount of hydrogen is 1.297*1.5 = 1.946 mole. At this point, you could use a more sophisticated model of gases to get the volume, if your course has covered those, or you could just treat it as an ideal gas with a molar volume of 22.4L, so the volume is 22.4*1.946 = 43.6 liters.
PV=nRT moles of aluminum = 35/ 26.98 = 1.2973 (3Mole H2l/ 2mole Al) = 1.94595 moles H2 (1atm)(V)= (1.94595) (.0821) (273.15) V= 43.6 L 6HCl + 2Al(s) = 2AlCl3 + 3H2
