How do you originate a word problem by using mass-mass when using mole?
600524
Give the mass of one reagent in a reaction, and ask for the mass of another. A word equation or chemical equation must be provided.\n\neg:\n\nWhat mass of nitrogen gas is required to produce 45.0 g of ammonia, according to the reaction below ?\n\n         N2   + 3 H2   ---> 2 NH3\n\n\nTo solve the problem\n\na) convert mass of NH3 to moles (2.65 mol)\nb) use stoichiometry - find moles of N2 needed (1.32 mol) \nc) convert moles of N2 to mass (37.0 g)
715558
OK, but you need to look up atomic weights or calculate molar mass.\n\n   Grams = Moles x molar mass  OR  moles x atomic mass\n\n    Moles = Grams/(molar mass) OR Grams/(atomic mass)
146781
mass = (side) x (side) x (side) or (height) x (depth) x (width)
264121
The theory can be used to calculate the mass of product that is expected to be produced when more than one reactant is involved in its formation.\nConsider the reaction below:\n\n2Na2S + Na2CO3 + 4SO2 ---> 3Na2S2O3 + CO2 \n\nGiven the mass of the reactants involved, we would use just one of them as a reference to get the mass of the product, but which reactant should we choose? \n\nHere we find the moles of each of the 3 reactants (given the mass)  and then compare the moles ratio of each in turn with the product and hence calculate the moles of the product Na2S2O3. A limiting reactant is one which gives the least moles of the product. So we see that from which reactant we get the least moles of Na2S2O3, and then with the moles, we can finally calculate the mass of product we expect to get when the reaction is carried out.\nWe choose the limiting reactant for finding out the yield because after any one of the reactants is used up,  the reaction won't proceed further, so this is to calculate the maximum amount of product we can get.
843034
1. Modify the ideal gas equation such that molecular mass (MW)and weight of gas appear.\nIdeal gas equation: PV = nRT\nsince n (moles) = weight / MW = g/MW\nso, your equation will become PV = gRT/MW\nGiven your values, you will end up:\nMW = gRT/ PV\n\n2. Plug in the values:\n\nMW = 3.06g * 0.08206 L-atm/mol-K * (51.4 + 273 K) / 0.970atm * 1.00L\nMW = 83.98 g/mol\n- this is the molecular mass of your gas.\n\n3. Determine the molecular formula of the gas.\n-since you have 14.37% H, your C is 85.63%\n-assume 100g gas so that you'll have 14.37g H and 85.63g C\n-get the number of moles:\nH: 14.37g / 1g/mol = 14.37 mol H\nC: 85.63g / 12g/mol = 7.13 mol C\n\n- get the mole ratio: that is, 1 mol C is to 2 mol H.\n- determine your multiplier such that MW result will be approximately 83.98g/mol. In this case, its 6. \n\nSo, your molecular formula is C6H12.\n\nTo check,\n\nC: 6 * 12g/mol = 72g/mol\nH: 12 * 1g/mol = 12g/mol\ntotal: 84g/mol ~ 83.98g/mol
727216
use beers law to figure out the concentration and then take that to mole ratio
783725
All right, you know what M means, right?  It's short for moles per liter.  You want to have 0.250 moles of Al2(SO4)3 in each liter of solution.  You're preparing 0.4 L of solution, so you need (0.4 * 0.250) = 0.1 moles of Al(SO4)3.\n\nNow the molar mass of each element is listed on most periodic tables.  That's how much one mole of that atom weighs.  A molecule weighs as much as all the atoms in it put together, so a mole of Al2(SO4)3 weighs as much as two moles of Al, 3 moles of S and 12 moles of O.  That's 2(26.982) + 3(32.06) + 12(15.999), according to my periodic table.  So if you add it all up, a mole of Al2(SO4)3 weighs 342.132 grams.  You only need a tenth of a mole, so you use 34.2132 grams.
681504
For this you need to the density or molarity of the solution.\n\nUsing molarity:\nMolarity=(Number of moles of solute)/(litre of solution)\n\nAlso 1 litre=1000 cubic centimeter.\n\nUsing density, d=m/V you get the mass of the solute.\nFrom there you can go to moles by using \nmoles=mass/(atomic or molecular mass)
569202
Scientists use it today to calculate the amount of a substance, number of particles, that they have.  The molecular mass of something times the number of moles of the substance equals the mass in grams of the substance.  For example 1 mol of H2O equals 18g, and equals 6.02810^23 particles of water.
459258
start with computation of molarity of solution:\nM_BaOH = (51.24g/ MW BAOH)/ 1.2 L\n\nthen \nusing M1V1 = M2V2\nM_BaOH (x) = (0.1) (1L)\nsolve for x
851431
MASS / VOLUME\n that's way you measure it in g/mL
313383
Molarity= moles/liters
765367
The mole (symbol: mol) is the SI base unit that measures an amount of substance. One mole contains Avogadro's number (approximately 6.022×1023) entities. A mole is much like a dozen in that both units can describe any set of elementary objects, although the mole's use is usually limited to measurement of subatomic, atomic, and molecular structures, as a mole represents an exceedingly large number of entities.
961444
We will need to find if there is a limiting material.\n\nCO + 2H2 ------> CH3OH\n\n(1.61 g H2)*(1 mole H2/2.016 g) = 0.7986 mole H2\n\n(0.7986 mole H2)*(0.5 mole CH3OH/mole H2) = 0.3993 mole CH3OH at most\n\n(15.95 g CO)*(1 mole CO/28g) = 0.5696 mole CO\n\n(0.5696 mole CO)*(1 mole CH3OH/1mole CO) = 0.5696 mole CH3OH at most\n\nH2 is the limiting reagent, therefore:\n0.3993 mole CH3OH * (32.032 g/mole CH3OH) = 12.79 g\n\nPlease check for errors, I went by memory on what the atomic and molecular weights were, you may want to check this.
290716
Unless a equation is properly balanced the amount of product and reactant will not be the same, I suppose this is like a recipe where u get the proper number of ingredients(moles) and mix to together to give the you the proper number of products.
994150
A mole is simply a way of saying theres a biiiiig number of atoms involved. Just think of it as say, the word "billion". I have a "billion" atoms. What is a billion? Well a billion is 1,000,000,000. What is a mole? Well a mole is 6.022*10^23. If theres 1.5 billion molecules in a sample, there are 1,5000,000,000 molecules in that sample. If there is 1 mole of molecules in a sample, there are 6.022*10^23 molecules in that sample. Simple as that, it's just a very large number.\n\nIf have you have a formula that says for example: 1 NaOH + 1 HCl -> 1 NaCl + 1 H20, you can look at it with moles in mind. For every 1 mole of Sodium Hydroxide and 1 mole of Hydrochloric acid, you will produce 1 mole of sodium chloride and1 mole of water. It basically means that if you add 6.022*10^23 molecules of sodium hydroxide to the same number of hydrochloric acid, you'll produce 6.022*10^23 molecules of sodium chloride and water.
921703
Just let the units tell you what to do.  H=mass* T*sp heat.  with T in K