The metal is 6 a 5.00 g piece of metal is heated to 100. Then places in a beaker containing 20.
The balanced equation for the reaction between KOH and HBr is as follows;
KOH + HBr --> KBr + H₂O
stoichiometry of KOH to HBr is 1:1
number of KOH moles reacted - 0.25 mol/L x 0.015 L = 0.00375 mol
according to molar ration
number of KOH moles reacted = number of HBr moles reacted
number of HBr moles reacted - 0.00375 mol
if 12 mL of HBr contains - 0.00375 mol
then 1000 mL of HBr contains - 0.00375 mol / 12 mL x 1000 mL = 0.313 mol
therefore molarity of HBr is 0.313 M
Answer:
= 100u. Hence 10 g = 0.1 mole. Hope it's helpful to u
Answer:
13.8 moles of water produced.
Explanation:
Given data:
Moles of KMnO₄ = 3.45 mol
Moles of water = ?
Solution:
Chemical equation:
16HCl + 2KMnO₄ → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O
Moler ratio of water and KMnO₄:
KMnO₄ : H₂O
2 : 8
3.45 : 8/2×3.45 = 13.8 mol
Hence, 3.45 moles of KMnO₄ will produced 13.8 mol of water.
