Presentation is loading. Please wait.

Presentation is loading. Please wait.

Stoichiometry .

Similar presentations


Presentation on theme: "Stoichiometry ."— Presentation transcript:

1 Stoichiometry

2 Question 18 What mass of NaCl will be produced by the reaction of 58.7g of NaI with 29.4g of Cl2 gas if the products are sodium chloride and I2? Go To Step One

3 Step One Choice A Choice B Choice C Balance The Equation.
NaI + Cl NaCl + I2 2NaI + Cl NaCl + I2 Choice C 2NaI + Cl NaCl + I2

4 Choice A NaI + Cl NaCl + I2 Hint: There has to be the same number of Iodide on each side. Try Again

5 Choice B 2NaI + Cl NaCl + I2 Hint: There has to be the same number of Na on each side. Try Again

6 Correct Choice 2NaI + Cl2 2NaCl + I2
Note: There is the same number of Na, I, and Cl on each side. Go To Step Two

7 Step Two Write out the equation in the correct form by starting off with the 58.7g of NaI. Then, find the molar mass of NaI. Multiply 58.7g of NaI by 1 mole of NaI and put the molar mass on bottom by looking at the periodic table. After that, multiply both of those terms by the number of moles of each element. Put the element you want to find on top,(NaCl) and the moles of the original element, (NaI) on bottom to cancel out. Multiply those terms by the molar mass of the element you are looking for and put 1 mole of NaCl on bottom to cancel out. Choice A

8 58.7g NaI x 1 mol NaI x 2 mol NaCl x 58.44g NaCl
149.9g NaI 1 mol NaI mol NaCl Choice A 58.7g NaI x 1 mol NaI x 2 mol NaI x 58.44g NaCl 149.9g NaI 2 mol NaCl 1 mol NaCl Choice B Choice C 58.7g NaI x I mol NaI x 2 mol NaCl x 58.44g NaCl 149.9 g NaI 2 mol NaI 1 mol NaCl

9 Choice A 58.7g NaI x 1 mol NaI x 2 mol NaCl x 58.44g NaCl
149.9g NaI 1 mol NaI mol NaCl Hint: The number of moles is supposed to be a higher number. Try Again

10 Choice B 58.7g NaI x 1 mol NaI x 2 mol NaI x 58.44g NaCl
149.9g NaI 2 mol NaCl 1 mol NaCl Hint: The moles of NaI has to cancel out on bottom. Try Again Try Again

11 Correct Choice 58.7g NaI x I mol NaI x 2 mol NaCl x 58.44g NaCl
149.9 g NaI 2 mol NaI 1 mol NaCl Notice that everything crosses out on bottom and on top. Go To Step Three

12 Step Three 58.7 x 1 x 2 x 58.44 = 22.9 gNaCl 149.89 x 2 x 1
Multiply everything on top and then divide it by everything on bottom and now you have your first answer. 58.7 x 1 x 2 x = gNaCl x 2 x 1 Go To Step Four

13 Step Four To find the second equation and answer, do the same thing you just did except you start with the 29.4g of Cl2 and you want to find the grams of NaCl. 29.4g Cl2 x 1 mol Cl2 x 2 mol NaCl x 58.44g NaCl 70.9g Cl mol Cl mol NaCl Go To Step Five

14 Step Five Multiply everything on top and then divide it by everything on bottom. This is your second answer. 29.4 x 1 x 2 x = 48.5g NaCl 70.9 x 1 x 1 Go To Step Six

15 Step Six Now compare your first answer with your second answer and see which one is smaller. Since 22.9g of NaCl is smaller, NaI is your limiting reactant because that is the reactant that you started with. 1) gNaCl 2) g NaCl

16 THE END 


Download ppt "Stoichiometry ."

Similar presentations


Ads by Google