MnO2(s) + 4HCl(aq) → MnCl2(aq) + Cl2(g) + H2O
How many grams of HCl are required to produce 11.2 liters of Cl2? The reaction occurs in the presence of excess HCl and at STP.
A) 20.0 g
B) 36.5 g
C) 48.5 g
D) 73.0 g
Answers
73 grams of HCl produces 11.2 litres of Cl₂
Option (d) is correct.
Explanation:
Volume of one mole of Cl₂ at STP = 22.4 litres
Therefore in 11.2 liters, moles of Cl₂ - 0.5
According to the given equation,
4 moles of HCl produces 1 mole of Cl₂
or, 4 × 36.46 grams of HCl produces 1 mole of Cl₂
or, 2 × 36.46 grams of HCl produces 0.5 moles of Cl₂
or, 72.92 grams of HCl produces 0.5 moles of Cl₂
Therefore, option (d) is correct.
Know More:
Q: When 22.4 litres of H2 (g) is mixed with 11.2 liters of Cl2 (g), each at STP, the moles of HCl(g) formed is equal to
(a) 1 mol of HCl (g)
(b) 2 mol of HCl (g)
(c).0.5 mol of HCl(g)
(d) 1.5 mol of HCl (g)
Click Here: https://brainly.in/question/1311386
Q: 4HCl(aq) + MnO2(s) - 2H2O(L) + MnCl2(aq) + Cl2(g).
How many grams of HCl would react with 5 grams on MnO2, if Mn=55u and Cl=35.5u ?
Click Here: https://brainly.in/question/1359257