(FINAL) Chemistry Notebook 2016-17

Moles to Mass: (*in a balanced equation, everything on both sides is equal to each other as a ratio)
(1) For the balanced equation shown below, how many grams of S reacted, if 0.147moles of Al^2S^3 are produced?
_2_Al+_3_S=>_1_Al^2S^3
1. Take the amount you are given (in this case, 0.147 moL Al^2S^3) and convert to moLs of the desired element
(through the balanced equation given)
0.147 _____________________________
moL Al^2S^3 // 3 moL S
1 moL Al^2S^3
2. Take the moLs of the substance you've converted to (3 moLs of S), and turn those into grams (the desired unit = grams of S)
___________________________________________
0.147 moL Al^2S^3 // 3 moL S // 32 g of S
1 moL Al^2S^3 // 1 moL of S
3. Multiply across the top and the bottom
___________________________________________ 0.147 moL Al^2S^3 // 3 moL S // 32 g of S
->
1 moL Al^2S^3 // 1 moL of S
14.112 ______
1
-> 14.112 g of S
4. Apply significant figures
14.112 g of S -> (only 3 significant figures given at the beginning of the problem) -> 14.1 g of S
The answer is : 14.1 g of S
Mass to Mass:
(*in a balanced equation, everything on both sides is equal to each other as a ratio)
(1) For the balanced equation shown below, how many grams of Al will react with 12.4 grams of Al2O3?
_2_Al+_3_MnO=>_1_Al^2O^3+_3_Mn
1. Take the amount you are given (in this case, 12.4g Al^2O^3) and convert to the moLs of the compound
^
(*find the molar mass of the compound so you can convert)
Al 2 * 27
O 3 * 16
^
amount of the element you have
54+48= 102g/moL Al^2O^3
12.4g __________________________
Al^2O^3 // 1 moL Al^2O^3
// 102 g Al^2O^3
2. Convert to the desired element/compound through mole:mole ratios (in this case, to moLs of Al through the balanced equation)
__________________________________________
12.4g Al^2O^3 // 1 moL Al^2O^3 // 2 moL Al
// 102 g Al^2O^3 // 1 moL Al^2O^3
3. Find the mass of the desired element through conversion of the mole
___________________________________________________
12.4g Al^2O^3 // 1 moL Al^2O^3 // 2 moL Al // 27g Al
// 102 g Al^2O^3 // 1 moL Al^2O^3 // 1 moL Al
4. Multiply across the top and the bottom, then divide.
___________________________________________________ 12.4g Al^2O^3 // 1 moL Al^2O^3 // 2 moL Al // 27g Al ->
// 102 g Al^2O^3 // 1 moL Al^2O^3 // 1 moL Al
_____ 669.6
102
-> 6.564705882 g Al -> (*apply sig. fig.) -> 6.56 g Al
107