Br2 + Cl2 ---> 2BrCl
1. FIND ALL THE REASONS ESTEQUIOMETRICAS WITH REGARD TO Cl2.
2. all the masses of Cl2 are needed to form 55g of BrCl.
3. all the grams of Br2 are needed to form 38g of BrCl.
please I need help for this question for a homework.
thanks
Look at the mole ratio: one mole of Br2 produces two moles of BrCl
So for 2) how many moles of BrCl is in 55grams? Figure that out. Then, you must have half that amount of moles of Cl2. Convert that to grams.
http://es.wikipedia.org/wiki/Estequiometr%C3%ADa
I NOT UNDERSTAND
1. To find the stoichiometric ratios with respect to Cl2 in the given chemical equation, we need to examine the coefficients in front of Cl2. In this case, the coefficient is 1. This means that for every 1 mole of Cl2, we will form 2 moles of BrCl.
2. To calculate the mass of Cl2 required to form 55g of BrCl, we need to use the molar mass of BrCl and the stoichiometric ratio between Cl2 and BrCl.
First, we need to determine the molar mass of BrCl by adding the atomic masses of Br (79.904 g/mol) and Cl (35.453 g/mol).
Molar mass of BrCl = 79.904 g/mol + 35.453 g/mol = 115.357 g/mol
Next, we can use the stoichiometric ratio of 1 mole of Cl2 to 2 moles of BrCl to set up a proportion:
(1 mole Cl2) / (2 moles BrCl) = (x g Cl2) / (55g BrCl)
Cross-multiply and solve for x:
x g Cl2 = (1 mole Cl2 / 2 moles BrCl) * (55g BrCl)
x g Cl2 = 27.5g Cl2
Therefore, 27.5 grams of Cl2 are needed to form 55g of BrCl.
3. To calculate the grams of Br2 required to form 38g of BrCl, we will use the molar mass of Br2 and the stoichiometric ratio between Br2 and BrCl.
The molar mass of Br2 is simply the atomic mass of Br (79.904 g/mol) multiplied by 2:
Molar mass of Br2 = 79.904 g/mol * 2 = 159.808 g/mol
Similar to the previous calculation, we can set up a proportion using the stoichiometric ratio of 1 mole of Br2 to 2 moles of BrCl:
(1 mole Br2) / (2 moles BrCl) = (x g Br2) / (38g BrCl)
Cross-multiply and solve for x:
x g Br2 = (1 mole Br2 / 2 moles BrCl) * (38g BrCl)
x g Br2 = 19g Br2
Therefore, 19 grams of Br2 are needed to form 38g of BrCl.