Sodium sulfate is not prepared from hydrogen chloride.
Let us find moles first. Molarity = moles of solute/Liters of solution ( 750 ml = 0.750 Liters ) 0.375 M Na2SO4 = moles Na2SO4/0.750 Liters = 0.28125 moles Na2SO4 =================== 0.28125 moles Na2SO4 (142.05 grams/1 mole Na2SO4) = 39.95 grams Na2SO4 needed ---------------------------------------you do significant figures!
na2so4= 142.05 g/mole, use dimensional analysis & set up your problem 13.64g
Na2SO4 10.0 grams Na2SO4 (1 mole Na2SO4/142.05 grams)(2 mole Na/1 mole Na2SO4)(22.99 grams/1 mole Na) = 3.24 grams of sodium -------------------------------
25.0 grams Na2SO4 ( only way this compound is possible ) 25.0 grams Na2SO4 (1 mole Na2SO4/142.05 grams) = 0.176 moles Na2SO4 -----------------------------
3.6 moles N2SO4 (142.05 grams/1 mole Na2SO4) = 511.38 grams Na2SO4 ==================( you do significant figures )
112,64 g of NaOH and 138,1 g of H2SO4
how man molecules are there in 450 grams of Na2SO4. the simple formula to determine of mole is NO OF MOL= GIVEN MASS IN gm/MOL:MASS OF COMP: , AND IMOL = 6.02X1023 . SO, 19. 077X1023 molecules are present in 450 grams of Na2SO4.
Balanced equation. 2NaOH + H2SO4 -> Na2SO4 + 2H2O 12.5 grams NaOH (1 mole NaOH/39.998 grams)(1 mole Na2SO4/2 mole NaOH)(142.05 grams/1 mole Na2SO4) = 22.2 grams sodium sulfate produced ===========================
The answer is 1,000 grams
A US pint is 473 grams. - An Imperial pint is 568 grams
Divide mass by molar mass to find moles:458(g) / 142.04(g/mol) = 3.22 mol Na2SO4
Full formal set up. 2.88 grams Na2SO4 (1 mole Na2SO4/142.05 grams)(6.022 X 1023/1 mole Na2SO4)(1 mole Na2SO4 atoms/6.022 X 1023) = 0.020 moles of sodium sulfate atoms ------------------------------------------------------( you can see the last two steps are superfluous )
Four and half
There are 0.5 moles in one litre of the solution, therefore 0.038 moles are present in... 0.076 litres answer: 76ml of Na2SO4 are needed.
how many grams of calcium nitrate are needed to make a 500ml volume of a .5 molar solution
Grams NaOH?? Balanced equation. 2NaOH + H2SO4 --> Na2SO4 + 2H2O 4.9 grams H2SO4 (1 mole H2SO4/98.086 grams)(2 mole NaOH/1 mole H2SO4)(39.998 grams/1 mole NaOH) = 4.0 grams NaOH needed =================
Balanced equation first. BaCl2 + Na2SO4 -> 2NaCl + BaSO4 22.6 ml BaCl2 = 0.0226 liters 54.6 ml Na2SO4 = 0.0546 liters 0.160 M BaCl2 = moles BaCl2/0.0226 liters = 0.00362 moles BaCl2 0.055 M Na2SO4 = moles Na2SO4/0.0546 liters = 0.0030 moles Na2SO4 The ratio of BaCl2 to Na2SO4 is one to one, so either mole count wull drive this reaction. Use 0.0003 moles Na2SO4 0.0030 moles Na2SO4 (1 mole BaSO4/1 mole Na2SO4)(233.37 grams/1 mole BaSO4) = 0.700 grams of BaCO4 produced