you start with 33.1g of Sodium Sulfate and 15.2g C calculate the max. mass of sodium sulfide dat can b produce
145956
MY ANSWER IS :\n\n18.2g  of Na2S is produced.
938236
Step one: find the moles of H2SO4.  (2.7 moles per liter)*0.025 liters.\n\nStep two: multiply by two because the reaction is \n2NaOH + H2SO4 ----> 2H2O  + Na2SO4 and so you need two NaOH molecules per molecule of H2SO4.\n\nStep three: multiply by the molecular mass of NaOH.\n\nThat will get you the answer.
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)
1051035
1-. NaCl + H2O Sodium Chloride ( table salt )\n2-. NaSO3 +2 H2O Sodium Sulfate ( or sulfite I always forget which is the 3 and which is the 4 )\n3-. NaC2H3O2 + H2O ( I'm guessing.  You will want to check your periodic table for the charges on this one )\n4-.NH4Cl + H2O.  Ammonium Chloride\n5-.NSO2 + 3H2O\n6-. not sure\n\nI presume you are working salt solutions.  Ammonia was never my strong suit. I'm positive about the first two.
372026
CuSO4·5H2O(s) → CuSO4(s) + 5H2O(g)
225870
Take 33g of the salt and dissolve it in 67 g of water to get 33% salt solution.\n\nIts not the same as 0.3 N solution
951188
4. How many grams of zinc chloride can be formed from the reaction of 7.96 grams of zinc with excess HCL?\n\nZn + 2HCl → ZnCl2 + H2\n\n1 mole Zn produces 1 mole ZnCl2\nTherefore 65.7g Zn produces (65.7 + 2 x 35.5)g ZnCl2\nie ~65.7g Zn produces ~136.7g ZnCl2\nSo 7.96g Zn produces ~7. 96 x 136.7/65.7g ZnCl2\nie ~ 16.6g ZnCl2\n\n5. How many kilograms of copper sulfide could be formed from the reaction of 1.80 mol of copper with excess sulfur.\n\nCu + S → CuS\n\nSop 1 mole Cu produces 1 mole CuS\nTherefore 1.8 mole Cu produces 1.8 mole CuS\n1 mol CuS weighs ~(63.5 + 32.1)g = 95.6g\nSo 1.8 mol CuS weighs ~ 1.8 x 95.6g ≈ 172.1g = 0.1721kg\n\n6. If Copper(one) sulfide is partially roasted in air (reaction with 02), copper(one) sulfite is formed. Subsequently upon heating, the copper sulfite thermally decomposes to copper (one) oxide and sulfur dioxide. Write balanced chemical equations for these 2 reactions\n\nCu2S + 3O2 → Cu2SO3\nCu2SO3 → Cu2O + SO2
956489
2 SO2 + O2 --> 2 SO3\n\n\nok, 64.00 g oxygen times the molar mass of 02( 32.00) to get mols of O2 then time the mole ratio of O2 to SO2 which is (2 mol SO2/2 Mol O2) and then u have the mols of SO2 times that by the molar mass of SO2 to get the mass of SO2\n\n64.00 x 32.00 x 2/2 x molar mass of SO2
819316
CuSO4+Fe = FeSO4+Cu\nCompute the number of atoms of CuSO4 in 7 gms of it, the number of atoms in 2gm of Fe. Choose the smaller number of the two. That is the number of atoms of Cu formed. Compute its weight.
424592
H2SO4 + 2NaOH ----> 2H2O + Na2SO4\n\nYOU MUST SEPARATE AND CANCEL THE COMMON IONS (AQUEOUS ONES)
781614
Na2SO4 (aq) + Ba(NO3)2 (aq) --> 2 NaNO3 (aq) + BaSO4 (s)\n\nThe moles of Barium = 5.00 L * 0.15 mol/L = 0.75 moles\n\nThis is also the moles of sulfate needed:\n\n0.75 mol / 0.36 mol/L = 2.08 L of Na2SO4
141058
sodium sulfte is soluble, but barium sulfate would form as a precipitate. Thus\n\nBaNO3 + Na2SO4 -> BaSO4 + 2 NaNO3, or simply\n\nBa(2+) + SO4(2-) -> BaSO4
1101513
Your question is not very clear, however I am attempting it.\n1:  1 mol of Na3PO4 will have 3 mol of Na and 1 mol of PO4 ions.\n2:  2.5 mol of Na2SO4 will have 5 mol of Na and 2.5 mol of SO4 ions.\n3: 3 mol of CaSO4 will have 3 mol of Ca and 3 mol of SO4 ions.\n4: 4 mol of NH4NO3 will have 4 mol of NH4 and 4 mol of NO3 ions.
751855
The first step is to balance the equation first, which yields\n\n2SO2 + O2 ---> 2SO3\n\nNow, find the total number of moles of 160 g O2:\n\n# of mole = 160/(16*2)\n= 5 moles.\n\nAccording to the balanced equation, the mole ratio between SO2 and O2 is 2:1. In other words, it takes twice amount (in moles) of SO2 to react with 1 unit of O2.\n\nSince we react 5 moles of O2, we need 10 moles of SO2.\n\nMass of SO2 = # mole * molecular weight\n= 10 * (32 + 16*2)\n= 640 g of SO2 is needed.
1029892
Na2S (aq) + 2H2O (l) = 2NaOH + H2S (g)
833745
first you have to balance the equation:\n2SO2+O2=2SO3\n\n1) 96gO2*(1moleO2/32g O2)= 3mole O2\n2) 3mole O2*(2moles SO3/1mole O2)=6mole SO3\n3) 6moles SO3*(80.1g SO3/1mole SO3)=480g SO3
1138619
2NaCl+H2SO4=Na2SO4+2HCl\n2NaBr+H2SO4=Na2SO4+2HBr\none mole of NaCl weighs 59 gm\none mole of NaBr weighs103gm\none mole of Na2SO4 weighs 142gm\nusing the above you can calculate