Question

How many liters of HCl(g) measured at STP can be produced from 4.00 g of Cl₂ and excess H₂ according to the following equation:
H₂(g) + Cl₂(g)  2HCl(g)

Answer 1

Answer:- V=2.53 L solution :- the reaction is given below H2(g) + Cl2(g) → 2HCl(g) No of moles = mass / molar mass Molar mass of Cl2 = 70.906 g/mol Moles of Cl2 = 4.00/70.906 mol 1 mol of Cl2 produces 2 mol of HCI 4.00/70.906 mol of Clz produces (2*4.00)/70.906 mol of HCI since we know that from ideal gas equation, PV = nRT so, V = nRT/P eqn 1 it is given that m=(2*4.00)/70.906 moles P=1.0 atm T= 273 K R=0.082057 Latm/mol K So from equation 1 V = (2*4.00*0.082057*273770.906*1.0 V = 2.53 L (answer have three significant figures) Please give me thumbs up