How many grams of the reactant in excess will remain after the reaction?
The first step is to write the formula of the reactants and the products. mercury(II) perchlorate = Hg(ClO4)2 sodium dichromate = Na2Cr2O7 This is a double replacement reaction. The products are sodium perchlorate and mercury II dichromate. sodium perchlorate = NaClO4 mercury II dichromate = HgCr2O7 This is the precipitate. Hg(ClO4)2 + Na2Cr2O7 → NaClO4 + HgCr2O7 To balance this equation, we need two sodium perchlorate. Hg(ClO4)2 + Na2Cr2O7 → 2 NaClO4 + HgCr2O7 According to the coefficients in this balanced equation, one mole of mercury(II) perchlorate reacts with one mole of sodium dichromate to produce two moles of sodium perchlorate and one mole of mercury II dichromate. The next step is to determine the mass of one mole of each reactant. For Hg(ClO4)2, mass = 200.59 + 2 * 35.453 + 8 * 15.999 = 399.488 grams For Na2Cr2O7, mass = 22.99 * 2 + 51.996 * 2 + 15.999 * 7 = 261.965 gram The next step is to determine the number moles of each reactant. For Hg(ClO4)2, n = 42.404 ÷ 399.488 This is approximately 0.106 mole. For Na2Cr2O7, n = 10.872 ÷ 261.965 This is approximately 0.042. Since the ratio is 1:1, we have excess mercury(II) perchlorate. This means the number of moles of mercury II dichromate is equal to the number of moles of sodium dichromate. HgCr2O7 mass of mole = 200.59 + 51.996 * 2 + 15.999 * 7 = 416.575 Mass = (10.872 ÷ 261.965) * 416.575 This is approximately 17.35 grams. I hope this is helpful for you.
How many grams of the excess reactant is left?
i'm no longer going to respond to the question, yet i visit allow you to appreciate what to do. a everyday rule in Chemistry says that when you get caught with something, do a mole calculation. Get the ratio of NH3 to O2 in moles, then artwork out how lots NH3 continues to be, and artwork that lower back to the burden. i could desire to remark: i won't be able to be waiting to think of the respond interior the 1st poster is quite. i may be incorrect, yet in case you began with in basic terms a million,85g, you heavily isn't waiting to ultimately finally end up with over 17g. this could mean NH3 is formed, and by way of no ability used. consequently, first answer is defective, and ought to no longer be used. EDIT: In respond to very final poster (Stephanie D), you heavily isn't waiting to actual in basic terms subtract the weights from one yet yet another, in basic terms as you heavily isn't waiting to subtract apples from pears. the burden of an O2 molecule and the burden of a NH3 molecule differs very much, and you heavily isn't waiting to below any situation in basic terms state that they react by way of technique of weight. Reactions take place in ratios of the quantity of molecules present day (e.g. one O atom will bind with 2 H atoms to variety H2O, consequently a million Mole O binds with 2 Moles H to variety a million Mole H2O. might desire to be seen right here that your technique won't artwork, as H2 has an atomic weight of two and O has an atomic weight of sixteen), and in basic terms from there are you waiting to artwork lower back to weights working with the molecular weight of each and every and each atom. PLEASE by way of no ability make that mistake lower back, on your very own reliable! something of your calculations seems reliable regardless of the shown fact that.
Which reactant is in excess and how many grams will remain after the reaction is complete?
(26.0 g CaCO3) / (100.0875 g CaCO3/mol) = 0.25977 mol CaCO3 (14.0 g HCl) / (36.4611 g HCl/mol) = 0.38397 mol HCl 0.38397 mole of HCl would react completely with 0.38397 x (1/2) = 0.191985 mole of CaCO3, but there is more CaCO3 present than that, so CaCO3 is in excess and HCl is the limiting reactant. (0.38397 mol HCl) x (1 mol CaCl2 / 2 mol HCl) x (110.9840 g CaCl2/mol) = 21.3 g CaCl2 ((0.25977 mol CaCO3 initially) - (0.191985 mol CaCO3 reacted)) x (100.0875 g CaCO3/mol) = 6.78 g CaCO3 left over
Chemistry help, please! How many grams of the reactant in excess will remain after the reaction?
Write a balanced equation of the reaction: Hg(ClO4)2 + Na2S → HgS + 2NaClO4 1mol Hg(ClO4)2 reacts with 1 mol Na2S Molar mass Hg(ClO4)2 = 399.4912 g/mol Molar mass Na2S = 78.0445 g/mol Mol Hg(ClO4)2 in 68.795g = 68.795 / 399.4912 = 0.1722 mol Mol Na2S in 12.026g Na2S = 12.026/78.0445 =0.1541 mol The Hg(ClO4)2 is in excess Mol in excess = 0.1722 - 0.1541 = 0.0181 mol Mass of 0.0181 mol Hg(ClO4)2 = 0.0181*399.4912 = 7.234g There is 7.2364g Hg(ClO4)2 remaining
Chemistry question help?!?
Hg(NO3)2 (aq) + Na2S (aq) ==> HgS (s) + 2 NaNO3 (aq) MW of: Hg(NO3)2: 200.59+2*14+6*16 = 324.59 g/mole Na2S: 2*23+32.06 = 78.06 g/mole HgS = 200.59+32.06 = 232.65 g/mole Moles Reacted: Hg(NO3)2: (88.445 g) / (324.59 g/mole) = 0.2725 moles Na2S: (13.180 g) / (78.06 g/mole) = 0.1688 moles The balanced reaction equation tell us that the Hg(NO3)2 and Na2S react in a 1:1 mole ratio, and that the solid product will be produced is a 1:1 mole ratio to both reactants. Since there are more moles of Hg(NO3)2 to start, it is in excess, and Na2S is the limiting reagent. We can expect no more than 0.1688 moles of product, assuming 100% conversion of the reaction. 1. (0.1688 moles) (232.65 g/mole) = 39.282 g of HgS produced. 2. (0.2725 moles - 0.1688 moles) (324.59 g/mole) = 33.66 g of Hg(NO3)2 left over.
Not necessarily. Imagine a simple reaction where A and B combine to give AB (I'm on mobile so I'm not going to write in all the proper notation, if someone would like to edit it into my answer please, but all means). A is some very light species, and B is a big heavy molecule. There're about 3x as many moles of B as there are of A in the reactants. This means A is the limiting reactant.If you take away half of the amount of B (plus an amount of B equal to half the mass of A, which should be very little B because A is very light), you still have roughly 1.5x more moles of B than A, so A is still limiting and you produce exactly the same amount of AB.Of course, this example is contrived, but the point is that you need to know more about your system to be able to accurately predict how much product you'll get.In reality things are even more complicated, and the amount of product you get will depend on reaction equilibria in addition to just limiting reactants. Maybe the reaction proceeds poorly at low pressures, so with exactly the same amount of reactants you can vary the amount produced wildly by putting the system in a pressure chamber. Maybe the reaction needs a catalyst. Maybe A is a gas you're binding to a solid B substrate, and the reaction depends on the available surface area of B as well as the flow conditions for A. There is just so much to consider.For some very simple reactions in solution where you have the exact stoichiometric ratio of reactants (or you halve the amount of the limiting reactant), under constant conditions, and where the reaction goes to completion, then yes, halving the reactant mass will halve the product mass.
Chemistry problems. Help please?
We'll have to assume that you mean 10 grams of hydrogen chloride in the first part. Can you convert that 10g to moles? and do the same with the 28g of calcium carbonate. then look at the equation; you need twice as many moles of HCl, compared to CaCO3 if you've got more than twice as much, then CaCO3 will be the limiting reactant. if you've got less than twice as much, HCl will be the limiting reactant. The moles of limiting reactant will determine the moles of product when you get the moles, you convert them back into grams by multiplying by the molar mass. do the same process with the sulphur equation: convrt masses to moles look at mole ratio in equation to decide which is the limiting reagent and which is in excess. find the moles of product. convert moles back into grams. once you've done it a few times it'll get easier.
Does the quantity of reactant have an impact on the reaction rate?There is a whole scientific discipline that is related to this question: Chemical Kinetics or reaction kinetics.For reactants in the gas phase or in solution, reaction rates are in general proportional to the concentrations of the reactants. In the case of the reaction of solid magnesium with diluted hydrochloric acid, the reaction rate depends on the surface area of the magnesium and the concentration of the acid.Rate of Reaction of Magnesium and Hydrochloric Acid