Match the column I with column II and mark the appropriate choiceColumn I Column II(i) Mass of H2 produced when 0.5 mole of zinc reacts with excess of HCl (a) 3.011 × 1023 molecules(ii) Mass of all atoms of a compound with formula C70H22 (b) 6.023 × 1023 molecules(iii) Number of molecules in 35.5 g of Cl2 (c) 1.43 × 10–21 g(iv) Number of molecules in 64 g of SO2 (d) 1g

If 6.5 g of zinc reacts with 5.0 g of HCl, according to the following reaction. Zn + 2HCl -> ZnCl2 + H2 a) Which substance is the limiting reactant? b) How many grams of the reactant remain unreacted? c) How many grams of hydrogen would be produced?

Zinc metal reacts with aqueous hydrochloric acid to produce zinc chloride and hydrogen gas.​     Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g) What volume of 3.55 M HCl is required to produce 25.5 L of hydrogen gas at a temperature of 298 K and a pressure of 1.20 atm?Group of answer choices1.25 L2.50 L0.705 L8.88 L

Zinc reacts with hydrochloric acid according to the following balanced chemical equation:Zn (s)  +  2 HCl(aq)  →   ZnCl2(aq)  +  H2(g)A student needs to react 0.278 mol Zn with hydrochloric acid. Calculate the minimum number of moles of HCl required, assuming all the zinc reacts?Enter your answer to three significant figures. Do not include units.

How many moles of zinc will react with 12 moles of HCl? Zn (s) + 2 HCl (aq) yields ZnCl2 (aq) + H2 (g)2 marks6122410

To solve this problem, we will follow these steps: a) Determine the limiting reactant: Calculate the moles of each reactant: Moles of Zn: 6.5 g / molar mass of Zn Moles of HCl: 5.0 g / molar mass of HCl Use the stoichiometry of the reaction to determine the moles of product formed from each reactant: Moles of ZnCl2 from Zn: moles of Zn / 1 (from the balanced equation) Moles of ZnCl2 from HCl: moles of HCl / 2 (from the balanced equation) The limiting reactant is the one that produces the least amount of product. In this case, HCl is the limiting reactant because it produces less ZnCl2 compared to Zn. b) Calculate the grams of the reactant that remain unreacted: Calculate the moles of HCl consumed in the reaction. Calculate the moles of HCl remaining by subtracting the moles consumed from the initial moles. Convert the remaining moles of HCl to grams. c) Calculate the grams of hydrogen gas produced: Determine the moles of H2 produced from the moles of HCl consumed. Convert the moles of H2 to grams using the molar mass of hydrogen gas. Given: Molar mass of Zn = 65.38 g/mol Molar mass of HCl = 36.46 g/mol Molar mass of H2 = 2.02 g/mol Let's perform the calculations: a) Limiting reactant: Moles of Zn: 6.5 g / 65.38 g/mol = 0.0994 mol Moles of HCl: 5.0 g / 36.46 g/mol = 0.137 mol Moles of ZnCl2 from Zn: 0.0994 mol Moles of ZnCl2 from HCl: 0.137 mol / 2 = 0.0685 mol HCl is the limiting reactant because it produces less ZnCl2. b) Remaining unreacted HCl: Initial moles of HCl = 0.137 mol Moles of HCl consumed = 0.137 mol - 0.0685 mol = 0.0685 mol Remaining moles of HCl = 0.137 mol - 0.0685 mol = 0.0685 mol Remaining grams of HCl = 0.0685 mol * 36.46 g/mol = 2.50 g c) Grams of hydrogen gas produced: Moles of H2 produced = 0.0685 mol (from the consumed HCl) Grams of H2 produced = 0.0685 mol * 2.02 g/mol = 0.1387 g Therefore: a) HCl is the limiting reactant. b) 2.50 grams of HCl remain unreacted. c) 0.1387 grams of hydrogen gas would be produced.