Answer:
O₂; KCl; 33.3
Explanation:
We are given the moles of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble all the data in one place.
2KCl + 3O₂ ⟶ 2KClO₃
n/mol: 100.0 100.0
1. Identify the limiting reactant
(a) Calculate the moles of KClO₃ that can be formed from each reactant
(i)From KCl
(ii) From O₂
O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.
KClO₃ is the excess reactant.
2. Moles of KCl left over
(a) Moles of KCl used
(b) Moles of KCl left over
n = 100.0 mol - 66.67 mol = 33.3 mol
Answer:
120g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
Sn + 2HF —> SnF2 + H2
Next, we shall determine the number of mole of HF needed to react with 3 moles of Sn.
From the balanced equation above, 1 mole of Sn reacted with 2 moles of HF.
Therefore, 3 moles of Sn will react with = 3 x 2 = 6 moles of HF.
Finally, we shall convert 6moles of HF to grams
This is illustrated below:
Number of mole of HF = 6moles
Molar Mass of HF = 1 + 19 = 20g/mol
Mass of HF =..?
Mass = number of mole x molar Mass
Mass of HF = 6 x 20
Mass of HF = 120g
Therefore, 120g of HF is needed to react with 3 moles of Sn
The products of chemical reactions often have completely different properties than the reactants, like viscosity, boiling and melting temperatures, etc.
That is because the atoms form new and different bonds to give the products.
