3 Cl2 + 6 KOH go to KClO3 + 5 KCl + 3 H2O how many moles of KClO3 form when 2 grams of KOH react completely
Convert gms to moles (divide given mass in grams by formula weight) and use equation ratio... f.wt.(KOH) = 56 g/mole
2gms KOH = (2/56)mole KOH = 0.036mole KOH
fm equation 6molesKOH => 1moleKClO3
then ... (6/0.036) = (1/x) => x = (0.036/6)mole KClO3 = 0.006 mole KClO3
To find the number of moles of KClO3 formed, we need to follow these steps:
1. Determine the molar mass of KOH:
- The molar mass of K (potassium) is 39.10 g/mol.
- The molar mass of O (oxygen) is 16.00 g/mol.
- The molar mass of H (hydrogen) is 1.01 g/mol.
Adding these together gives us:
Molar mass of KOH = (39.10 g/mol) + (16.00 g/mol) + (1.01 g/mol) = 56.11 g/mol.
2. Calculate the number of moles of KOH:
- Given mass of KOH = 2 grams.
Number of moles = mass / molar mass
Number of moles of KOH = 2 g / 56.11 g/mol = 0.0356 mol.
3. Determine the stoichiometric ratio between KOH and KClO3:
According to the balanced equation, the stoichiometric ratio between KOH and KClO3 is 6:1.
4. Calculate the number of moles of KClO3 using the stoichiometric ratio:
Number of moles of KClO3 = (0.0356 mol KOH) * (1 mol KClO3/6 mol KOH)
= 0.00593 mol KClO3.
Therefore, when 2 grams of KOH react completely, approximately 0.00593 moles of KClO3 will be formed.