This ratio is 1:2.
The molar mass of CuSO4 is 159.6 g/mol. To find the mass of 3 moles of CuSO4, multiply the molar mass by the number of moles: 3 moles * 159.6 g/mol = 478.8 grams. Therefore, there are 478.8 grams in 3 moles of CuSO4.
1st Get the balanced equation NaOH + HCl -> NaCl + H2O Find the number of moles in HCl; n = cv n = 0.46x0.61 n = 0.2806 moles the number of moles of HCl and NaOH is the same so 0.2806moles will be needed
To neutralize HCl with NaOH, the mole ratio is 1:1. So, the moles of HCl are 0.200 M x 0.020 L = 0.004 moles. Since NaOH and HCl react in a 1:1 ratio, we need 0.004 moles of NaOH. Using the molarity formula, we find that we need 0.010 L or 10.00 mL of 0.400 M NaOH.
Well, darling, if we're talking about a 1:2 molar ratio between NaOH and H2SO4, then you'd need 2 moles of NaOH to neutralize 1 mole of H2SO4. It's all about those stoichiometry dance moves, honey. Just make sure you're not tripping over your chemical equations!
To determine the maximum volume of 0.10 M NaOH that can be neutralized by 0.20 Ml of HCl, you need to use the equation: moles = Molarity × Volume. First, calculate the moles of HCl used (0.20 ml * 0.20 M) and then use the mole ratio from the balanced chemical equation to determine the moles of NaOH needed. Finally, divide the moles of NaOH by the concentration of NaOH to find the volume that can be neutralized.
The ratio of moles of CuSO4 to moles of water in CuSO4•5H2O is 1:5. This is because there is one mole of CuSO4 for every five moles of water in the compound.
To determine the limiting reagent between CuSO₄ and NaOH, we first need to look at the balanced chemical equation for the reaction, which is: [ CuSO₄ + 2 NaOH \rightarrow Cu(OH)₂ + Na₂SO₄ ] According to the stoichiometry, 1 mole of CuSO₄ reacts with 2 moles of NaOH. First, calculate the moles of CuSO₄ and NaOH. Assuming the molar mass of CuSO₄ is approximately 159.61 g/mol, 638.44 g of CuSO₄ corresponds to about 4.00 moles. For 240.0 g of NaOH (molar mass ≈ 40.00 g/mol), this is about 6.00 moles. Since 4.00 moles of CuSO₄ would require 8.00 moles of NaOH but only 6.00 moles are available, NaOH is the limiting reagent.
moles of what?
The molar mass of CuSO4 is 159.6 g/mol. To find the mass of 3 moles of CuSO4, multiply the molar mass by the number of moles: 3 moles * 159.6 g/mol = 478.8 grams. Therefore, there are 478.8 grams in 3 moles of CuSO4.
To determine the grams of KHP required, you first need to calculate the number of moles of NaOH present in the 50 mL solution. Then, using the balanced chemical equation of the titration between NaOH and KHP, you can find the mole ratio. From the mole ratio and the moles of NaOH, you can calculate the moles of KHP needed and then convert that to grams of KHP.
Molarity = moles of solute/Liters of solution 3.42 M NaOH = 1.3 moles NaOH/Liters NaOH Liters NaOH = 1.3 moles NaOH/3.42 M NaOH = 0.38 Liters
1st Get the balanced equation NaOH + HCl -> NaCl + H2O Find the number of moles in HCl; n = cv n = 0.46x0.61 n = 0.2806 moles the number of moles of HCl and NaOH is the same so 0.2806moles will be needed
To neutralize HCl with NaOH, the mole ratio is 1:1. So, the moles of HCl are 0.200 M x 0.020 L = 0.004 moles. Since NaOH and HCl react in a 1:1 ratio, we need 0.004 moles of NaOH. Using the molarity formula, we find that we need 0.010 L or 10.00 mL of 0.400 M NaOH.
moles = mass/Mr moles = 100/(23+16+1) moles of NaOH = 2.5mol
3.42 moles NaOH (39.998 grams/1 mole NaOH) = 137 grams NaOH
Well, darling, if we're talking about a 1:2 molar ratio between NaOH and H2SO4, then you'd need 2 moles of NaOH to neutralize 1 mole of H2SO4. It's all about those stoichiometry dance moves, honey. Just make sure you're not tripping over your chemical equations!
I assume you mean 32.0 grams of NaOH and 450 milliliters of NaOH. Molarity = moles of solute/Liters of solution ( 450 ml = 0.450 liters ) get moles of NaOH 32.0 grams NaOH (1 mole NaOH/39.998 grams) = 0.800 moles NaOH Molarity = 0.800 moles NaOH/0.450 liters = 1.78 Molar NaOH