Question

in the reaction assume that you added 2.43 g of Mg (s) to the reaction flask and how the reaction goes fully to completion

Mg(s) + 2HCl(aq) + MgCl(aq) + H2(B) In the above reaction, assume that you added 2.43 grams of Mg(s) to the reaction flask and the reaction goes fully to completion. a. What is the theoretical weight in grams of MgCl(s) that can be produced ? (2.5 Points) (Show calculations) b. What is the weight in grams of HCl(aq) that will be required to react with Mg(s)? (2.5 Points) (Show calculations) c. How many moles of H2(g) will be produced ? (2.5 Points) (Show calculations) d. If H() is collected under standard temperature and pressure conditions (aka STP) where one mole of a gas occupies 22.4 Liters, what volume of H2(g) will be produced ? (2.5 Points) (Show calculations)

Answer 1

Giveng Mg(s) + 2Hce (aq) MgCl₂ (q) + H₂ Cg) mas of Mg = 2.43 g solutions can Molar may of Mg = 24g/mol Molar mass of He= 1+35.5 36.5 g/mol Molar mass of Mgu = 24+243515 - 95 gl mal Molar mas of H₂ = Qg1mol Mg (3) + 2HCl(aq) - MgCl₂ (aq) +H₂(g) 2ug 2x36.5 asglmol 291mol. - 73.0 24+ 71 gl mol When 24 g of rg is taken, MgCl₂ produced is 95gimo) 1 g of rg is taken, mga produced is When 95 24 When 2.43g of rg is taken, Molla produced is 95 x 2.43 So, theortical mass of 219.618g) rg la obtained is [9.618 g)

24 - Nour, Mg + 2Hce MgCl₂ + H₂ 24g of rg produces 739 of Hee 13 414 Producas 13 9 но 2.439 of mg produces 73 x 2.43 g of ree 17.391 g of HQ] © Mg + 2Hce MgCl₂ + H₂ Given maus of Mg = 2,43g. Molar may of Mg = 24 glmal No. of moles of Mg: - Molar mas 2.43. 0.10125 mol 24 Now, from reaction, 1 mol of rg prodrices I mo) of H₂ Nou, 8, 10125 md of rg produces 10.10125 mol ) Given mass - W 1 mol of Hy gas occupies 224 L of volume 0.10125 mol of the gas occupies 22.4x0.10125 2.268 L & Volume So, H₂ Volume occupied = 12.268L]