7c

Law of Conservation of Mass

in an ordinary chemical reaction - matter cannot be created or destroyed - no change in total mass occur - the mass of the product is equal to the mass of reactant 2Ag(215.8 g) + S(32.1 g) = Ag2S (247.9 g) - 2Ag (s) + S(s) -> Ag2S(s) - interpret as 2 moles of Ag + 1 mole of S = 1 mole of Ag2S

How many moles of Fe2O3 can be produced from 6.0 moles of O2

- given 4Fe(s) + 3O2 (g) -> 2FeO3 (s) - we can look at as 4 moles of Fe + 3 moles of 02 = 2 moles of Fe2O3 - given 6.0 moles of O2 and we need to find how many moles of Fe2O3 can be produced 1, make sure the equation is balanced, yes reactant side: Fe=4, O=6 product side: Fe=4, O=6 2, find our mole-mole factor that will connect us to O2 and Fe2O3 > 3 moles O2/ 2 moles Fe2O3 or vice versa 2, use the mole-mole factor to determine the moles of Fe2O3, we are given 6.0 moles of O2 so use: 2 moles of Fe2O3/ 3 moles of O2 3, do the calculation > 6.0 moles of O2 x 2 moles of Fe2O3/ 3 moles of O2 = 4.0 moles of Fe2O3

principals of mass calculation for chemical reactions

- 4Fe(s) + 3O2(g) -> 2Fe2O3(s) > If we are given a problem where we are given a mass of Fe203 and we need to calculate the mass of O2 we have to use the mole ratio, 3 moles of O2/ 2 moles of Fe2O3 > If we are given mass of Fe203 and we need to calculate mass of Fe use mole ratio, 4 moles of Fe/ 2 moles of Fe2O3 > If we are given the mass of FE And we need to calculate the mass of O2, Then we use mole ratio: 3 moles of O2/ 4 moles of Fe

mole-mole factors from an equation

- consider the following balanced equation 2Fe(s) + 3S(s) -> Fe2S3 (s) - we can read it as > 2 moles of Fe + 3 moles of S -> 1 mole of Fe2S3 - because of this balanced equation we can write a relationship based on the coefficent with a mole-mole factor > two reactants can be written as: 2 moles of Fe/ 3 moles of S or vice versa > product and reactant or product with a element inside (subscript = the moles that element has) can be written as: 2 moles of Fe/ 1 mole Fe2S3 or vice versa and 3 moles of S/ 1 mole of Fe2S3 and vice versa

how many grams of O2 are needed to produce 45.8 grams of Fe2O3 in the following reaction

- given 45 grams of Fe2O3 and we need to find how much grams are in O2. - grams of Fe2O3 -> MM of Fe2O3 -> mole-mole of Fe2O3 and O2 -> MM of O2 1, find the MM of Fe2O3 > Find each element atomic mass and put it in a conversion factor, remember that MM is only 1 mole so we ignore the coefficient and we also know that the subscript tell us how much moles are in that element of that particular compound, but we will worry about that later. - 1 mole of Fe2/ 55.85 g of Fe - 1 mole of O3/ 16.00 g of O > now that we have our conversion factors, determine how much moles are in each element using the subscripts - since Fe2 has a subscript of 2 it has 2 moles - since O3 has a subscript of 3 it has 3 moles > now multiply the amount of moles to its conversion factor, make sure everything cancels - 2 moles of Fe2 x 55.85 g of Fe2/ 1 mole of Fe2 = 111.7 g of Fe2 - 3 moles of O3 x 16.00 g of O3/ 1 mole of O3 = 48.00 g of O3 > now add them all up, 111.7 g of Fe2 + 48.00 g of O3 = 159.7 g of Fe2O3 > Molar Mass for Fe2O3 is 1 mole of Fe203/ 159.7 g of Fe203 (remember MM is just 1 mole of that compound even though the coefficient has 2) 2, now find the mole-mole conversion factor for O2 and Fe2O3 > O3 has a coefficient of 3, there it has 3 moles, so its 3 moles of O2 > Fe2O3 has a coefficient of 2, so it has 2 moles, so its 2 moles of Fe2O3 > make the conversion factor, 3 moles of O2/ 2 moles of Fe2O3 or vice versa, 3, Now find the MM of O2, > Find O2 atomic mass and put it in a conversion factor, remember that MM is only 1 mole so we ignore the coefficient and we also know that the subscript tell us how much moles are in that particular element, but we will worry about that later. > 1 mole of O2/ 16.00 g of O2 > now that we have our conversion factors, determine how much moles are in each element using the subscripts > since O2 has a subscript of two, then there are 2 moles of O2 > now multiply that mole to our MM conversion factor > 2 moles of O2 x 16.00 g of O2/ 1 mole of O2 = 32.00 g of O2 > our MM conversion factor is: 1 mole of O2/ 32.00 g of O2 or vice versa (remember MM is just 1 mole of that compound even though the coefficient has 3) 4, plug everything in, we were given 45.8 grams and we have to find grams of O2, so: 45.8 grams of Fe2O3 x 1 mole of Fe203/ 159.7 g of Fe203 x 3 moles of O2/ 2 moles of Fe2O3 x 32.00 g of O2/1 mole of O2/ = 13.8 g of O2

mass calculation for chemical reaction (this will be given very often than not)

- given a balanced equation, convert mass of a substance A to mass of a substance B using these steps: 1, convert mass of substance A to moles using molar mass of a A 2, convert mole of substance A to moles of substance B using the mole-mole ration of B and A in the balanced equation 3, Convert moles of Substance B to grams using the molar mass of B

consider the following balanced equation 3H2(g) + N2 (g) -> 2NH3 (g)

- make sure the equation is balanced, (reactant side) H= 6, N= 2 and on the product side N=2 and H=6 1, a mole-mole factor for H2 and N2 is: 3 moles of H2/ 1 mole of N2 2, a mole-mole factor for NH3 and H2 is: 2 moles of NH3/ 3 moles of H2

important to note

- when doing mole to mole, coefficients are the moles of that compound or element (4Fe(s) = 4 moles of Fe) and we also ignore the subscripts - when doing molar mass, we ignore the coefficient (Because molar mass only pertains to one mole of the compound) and so we just focus on the subscripts which tells us how much moles they're for a specific element in that compound which then helps us to find our mass - Subscripts are for molar mass - Coefficient are for Mole to mole

how many moles of Fe are needed to react with 12.0 moles of O2

1, make sure the equation is balanced, yes reactant side: Fe=4, O=6 product side: Fe=4, O=6 2, find our mole-mole factor that will connect us to Fe and O3 > 4 moles Fe/ 3 moles O2 or vice versa 2, use the mole-mole factor to determine the moles of Fe are needed to react, we are given 12.0 moles of O2 so use: 4 moles of Fe/ 3 moles of O2 3, do the calculation > 12.0 moles of O2 x 4 moles of Fe/ 3 moles of O2 = 16.0 moles of Fe (from 3 sigfigs)

we want to determine the mass of (grams) of NH3 that can be produced from 32 grams of N2 in the following equation. N2(g) + 3H2(g) -> 2NH3 (g)

1, we are given 32 g of N2 and we want grams of NH3 by using our equation. 2, so grams of N2 -> molar mass of N2 -> moles of N2 -> mole-mole factor NH3/N2 -> moles NH3 -> molar mass NH3 -> grams of NH3 3, find the molar mass of N2 > 1 mole of N/ 14.01 g of N > multiple the moles of N2 to the conversion factor, the subscripts represent how much moles they're, so N has 2 moles > 2 moles of N x 14.01 g of N/ 1 mole of N2 = 28.02 g of N2 > new conversion factor for MM of N2 is: 1 mole of N2/ 28.02 g of N2 4, find the mole-mole factor for NH3 and N2 from our equation (coefficient represents our moles) - NH3 has a coefficient of 2 so there is 2 moles of NH3 - N2 has no coefficient so its just one mole of N2 > 1 mole of N2/ 2 moles of NH3 or vice versa 5, find the molar mass for NH3 (remember the molar mass is only with 1 mole of NH3, so we leave the coefficient out and for the subscripts we add them later). > molar mass for N, 1 mole of N/ 14.01 g of N or vice versa > molar mass of H, 1 mole of H/ 1.008 g of H or vice versa 6, Now we leave our coefficient out because we are only doing 1 mole of NH3 and our subscript for each element of our compound represents how much moles each element has, so for NH3 > N has 1 moles in NH3 > H has 3 moles in NH3 (3(subscript) = 3 moles) - now find the molar mass for N using its moles and molar mass conversion, 1 moles of N x 14.01 g of N/ 1 mole of N = 14.01 g of N - now find the molar mass for H using its moles and molar mass conversion, 3 moles of H x 1.008 g of H/ 1 mole of H = 3.024 g of H - add them all together > 14.01 g of N > + > 3.024 g of H > = 17.03 g of NH3 7, our new molar mass for NH3, > 1 mole of NH3/ 17.03 g of NH3 8, put it all together and makes sure everything cancels, so 32 g of N2 x 1 mole of N2/ 28.02 g of N2 x 2 moles of NH3/ 1 moles of N2 x 17.03 g of NH3/ 1 mole of NH3/ = 39 g of NH3 (round to 2 sigfigs)

- converting mass to moles - a box of table salt, NaCl, contains 737 g of NaCl. How many moles of NaCl are in the box?

1, we are given 737 g of NaCl and we need moles of NaCl 2, find the molar mass of NaCl to connect moles to g, remember that molar mass is only 1 mole, so don't worry about the subscripts until later. > 1 mole of Na/ 22.99 g of Na and vice versa > 1 mole of Cl/ 35.45 g of Cl and vice versa 3, now multiply the moles that each element has in NaCl, the subscript represent the amount of moles. > Na, Na has 1 mole in NaCl, so therefore its 1 mole of Na x 22.99 g of Na/ 1 mole of Na = 22.99 g of Na > Cl, Cl has 1 mole in NaCl, so therefore its 1 mole of Cl x 35.45 g of Cl/ 1 mole of Cl = 35.45 g of Cl 4, now add up our two values, > 22.99 g of Na > 35.45 g of Cl > = 58.44 g of NaCl (rounded to the least dp) 5, make the conversion factor > 1 mole of NaCl/ 58.44 g of NaCl or vice versa 6, we see that we are given 737 g of NaCl and we want moles of NaCl, so use the conversion factor and the given > 737 g NaCl x 1 mole of NaCl/ 58.44 g of NaCl = 12.61 moles of Cl (round to the least sigfig which is 4)