Mole to Mole, Mole to Gram

Date 5/1/13

Mole to Mole, Mole to GramBy: Keenan Reardon.

Wednesday, May 1, 13

Question 1 Mole to Mole

✤ Start with a balanced equation

✤ N2 + 3 H2 ---> 2 NH3


Question 1

✤ if we have 2.00 mol of N2 reacting, how many moles of NH3 will be produced?


Process 1

✤ Look at the ratios

✤ N2 + 3 H2 ---> 2 NH3

✤ 2.00 mol of N2-->?molNH3


Answer 2

✤ In the balanced equation for every mole ofnitrogen there were 2 NH3’s so if we have 2 moles of nitrogen then we would need 4 molNH3

✤ =4mol NH3



✤ Suppose 6.00 mol of H2 reacted. How many moles of ammonia(NH3) would be produced?


Answer 2

✤ This again is about ratios.

✤ N2 + 3 H2 ---> 2 NH3

✤ The Hydrogen ammonia ratio is 3/2

✤ Since we have 6 moles of hydrogen then we would need 4 moles of NH3

✤ =4molNH3



✤ We want to produce 2.75 mol of NH3. How many moles of nitrogen would be required?


Process

✤ This is tricky so we start with what is given to us.

✤ Which is 2.75 mol of NH3--> ?MolN2

✤ We then would have to use the coefficients and cross multiply, leaving what we want to be left with on top.


Answer 3

✤ The ratio of the equation is 2/1

✤ We would then cross multiply 2.75/x =2/1

✤ =1.38 mol N2


Question 1 Mole to Mass

✤ Start with balanced equation.

✤ 2 KClO3 ---> 2 KCl + 3 O2


Questions 1

✤ 1.50 mol of KClO3 decomposes. How many grams of O2 will be produced?


Process

✤ 2 KClO3 ---> 2 KCl + 3 O2

✤ Find ratios, KClO3 and O2 is 2/3

✤ 1.50 mol of KClO3--> ?grams O2

✤ Cross multiply - 1.50/x times 2/3

✤ =2.25mol O2

✤


Answer 1

✤ Now that we have 2.25mol O2 we convert it to grams.

✤ We then multiply 2.25mol O2 by the molar mass of one mole of O2

✤ 2.25mol O2 X 32g O2

✤ =72gO2


Question 2

✤ We want to produce 2.75 mol of KCl. How many grams of KClO3 would be required?

✤ 2 KClO3 ---> 2 KCl + 3 O2


Process

✤ Ratios

✤ 2/2

✤ So we know that we are gonna use the same amount of moles, all we have to do now is convert it to grams

✤ 2 KClO3 ---> 2 KCl + 3 O2


Answer

✤ 2.75mol KClO3 X 122.55g KCLO3

✤ =337g KCLO3

✤ 2 KClO3 ---> 2 KCl + 3 O2


Question 3

✤ If 80.0 grams of O2 was produced, how many moles of KClO3 decomposed?

✤ Different question, but same process, only difference if it is backwards.

✤ 2 KClO3 ---> 2 KCl + 3 O2


Process

✤ Find Ratio

✤ O2 to KCLO3 is 3/2

✤ we then divide 80.0 grams of O2 by the molar mass of O2

✤ 80/32 = 2.50molO2

✤ 2 KClO3 ---> 2 KCl + 3 O2


Answer 3

✤ Now we cross multiply 2.50/x = 3/2

✤ This equals 1.67 molKClO3

✤ 2 KClO3 ---> 2 KCl + 3 O2 ✤ 80/32 = 2.50molO2


Mole to Mole, Mole to Gram

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