112,64 g of NaOH and 138,1 g of H2SO4
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 ===========================
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 =================
The balanced chemical equation for the reaction is: H2SO4 + 2NaOH -> Na2SO4 + 2H2O. To find the amount of Na2SO4 produced, first find the limiting reactant by calculating the moles of each reactant. Then, use the mole ratio from the balanced equation to determine the moles of Na2SO4 produced. Finally, convert moles to grams using the molar mass of Na2SO4 to find the final amount.
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.
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 -------------------------------
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 ===========================
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 =================
About 245 grams.
About 245 grams.
The balanced chemical equation for the reaction is: H2SO4 + 2NaOH -> Na2SO4 + 2H2O. To find the amount of Na2SO4 produced, first find the limiting reactant by calculating the moles of each reactant. Then, use the mole ratio from the balanced equation to determine the moles of Na2SO4 produced. Finally, convert moles to grams using the molar mass of Na2SO4 to find the final amount.
Sodium sulfate is not prepared from hydrogen chloride.
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.
The balanced chemical equation for the reaction is 1 mol of sulfuric acid reacts with 2 mol of ammonium hydroxide. Therefore, for 8 mol of ammonium hydroxide, 4 mol of sulfuric acid are needed. To calculate the grams of sulfuric acid needed, you would multiply the number of moles by the molar mass of sulfuric acid.
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 -------------------------------
3.6 moles N2SO4 (142.05 grams/1 mole Na2SO4) = 511.38 grams Na2SO4 ==================( you do significant figures )
For every mole of sodium hydroxide, you need 1 mole of sulfuric acid for neutralization. The molar mass of sodium hydroxide (NaOH) is 40.0 g/mol and sulfuric acid (H2SO4) is 98.1 g/mol. So, to neutralize 40 g of NaOH (1 mole), you would need 98.1 g of H2SO4 (1 mole). Therefore, to neutralize 10.0 g of NaOH, you would need 24.53 g of H2SO4.
First write down the BALANCED reaction equation. 2NaOH + H2SO4 = Na2SO4 + 2H2O Note the molar ratios are 2:1::1:2 So two moles of NaOH produces one mole of Na2SO4 Next calculate tghe moles of NaOH mol(NaOH) = 200/(23 + 16 + 1) = 200/40 = 5 mol(NaOH) = 5 (This figure is equivalent to '2' above) mol(Na2SO4) = 5/2 = 2.5 ( Equivalent to '1' above) mol(Na2SO4) ; 2.5 = mass(g) / (2 x 23) + 32 + (4 x 16)) 2.5 = mass(g) / 142) mass*Na2SO4) = 2.5 x 142 = 355 g