To find the number of moles of Na2SO4 in 25.0 g of the compound, you need to convert the mass to moles. First, determine the molar mass of Na2SO4, then divide the given mass by the molar mass to obtain the number of moles.
3.6 moles N2SO4 (142.05 grams/1 mole Na2SO4) = 511.38 grams Na2SO4 ==================( you do significant figures )
First.Get moles sodium sulfate.5.35 grams Na2SO4 (1 mole Na2SO4/142.05 grams)= 0.0377 moles Na2SO4-------------------------------------Second.Molarity = moles of solute/Liters of solution ( 330 mL = 0.33 Liters )Molarity = 0.0377 moles Na2SO4/0.33 Liters= 0.114 M Na2SO4=============
To calculate the number of moles of atoms in 2.88 g of NaSO, you first need to find the molar mass of the compound. Sodium sulfate (Na2SO4) has a molar mass of 142.04 g/mol. Then, you can use the formula n = m/M, where n is the number of moles, m is the mass, and M is the molar mass, to calculate the number of moles in 2.88 g of Na2SO4.
A mole of oxygen atoms has a mass of approximately 16 grams. A mole of O2 has a mass of approximately 32 grams. A mole is 6.02 x 1023 particles and as such a mole of oxygen atoms has only half the mass of a mole of oxygen molecules.
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 ===========================
To find the number of moles of Na2SO4 in 25.0 g of the compound, you need to convert the mass to moles. First, determine the molar mass of Na2SO4, then divide the given mass by the molar mass to obtain the number of moles.
To determine the number of moles in Na2SO4, you first need to know the molar mass of the compound. The molar mass of Na2SO4 (sodium sulfate) is calculated by adding the atomic masses of each element present: 2(Na) + 1(S) + 4(O) = 2(22.99 g/mol) + 1(32.07 g/mol) + 4(16.00 g/mol) = 142.04 g/mol. To find the number of moles, you divide the given mass of Na2SO4 by its molar mass.
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
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 -------------------------------
To find the grams of sodium in 0.820 moles of Na2SO4, first calculate the molar mass of Na2SO4: 2(Na) + 1(S) + 4(O) = 2(23) + 32 + 4(16) = 142 g/mol. Since each mole of Na2SO4 contains 2 moles of Na atoms, the molar mass of Na in Na2SO4 is 46 g/mol. Therefore, in 0.820 moles of Na2SO4, there are 0.820 moles * 2 moles Na * 46 g/mol = 75.32 grams of sodium.
The mole ratio of salt (Na2SO4) to water in Na2SO4.10H2O is 1:10. This means that for every 1 mole of Na2SO4, there are 10 moles of water molecules.
The mole ratio for the given equation is 1:2:1:1. This means for every 1 mole of MgSO4, we need 2 moles of NaCl to react and produce 1 mole of Na2SO4 and 1 mole of MgCl2.
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.
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 )
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