Can you please check my answer?
Balance the following equation:
Cl2 + KI --> I2 + KCl
How many grams of Cl2 are used to form 6.8 g of KCl?
So 6.8 grams of KCl divided by 74.55g Kcl x 1molCl2 divided by 2mol KCl multiplied by 70.91 g Cl2/mol....
3.2 grams of Cl2
Cl2 + 2KI --> I2 + 2KCl
6.8gKCl*1molKCl/74.55gKCl*1molCL2/2MolKCl* 70gCL2/1molCl2
6.8*1*70/(74.55*2*1)
so you did it correctly.
thank you
To balance the given equation Cl2 + KI --> I2 + KCl, we need to ensure that the number of atoms of each element is the same on both sides of the equation.
The balanced equation is: Cl2 + 2KI --> I2 + 2KCl
Now, let's calculate the number of moles of KCl using the given mass and the molar mass of KCl.
6.8 g KCl / 74.55 g/mol KCl = 0.0912 mol KCl
From the balanced equation, we can see that 2 moles of KCl react with 1 mole of Cl2.
Next, we need to convert moles of KCl to moles of Cl2 using the stoichiometric ratio.
0.0912 mol KCl x (1 mol Cl2 / 2 mol KCl) = 0.0456 mol Cl2
Finally, we can convert moles of Cl2 to grams using the molar mass of Cl2.
0.0456 mol Cl2 x 70.91 g/mol Cl2 = 3.24 g Cl2
Therefore, the correct answer is approximately 3.24 grams of Cl2 are used to form 6.8 g of KCl.