Simple equality.
(X L)(2.5 M HCl) = (1.5 L)(5.0 M NaOH)
2.5X = 7.5
X = 3.0 Liters needed
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One mole to react with ONE mole CH3COOH (ethaancarbonic acid) because it has only ONE proton (H+) to react with, the other 3 Hydrogen's are not proteolytic (=NON-acidic)
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.
The balanced chemical equation for the reaction is: 3 H3PO4 + 2 NaOH -> Na3PO4 + 3 H2O From the equation, we can see that 3 moles of H3PO4 react with 2 moles of NaOH. Calculate the molar mass of H3PO4 and NaOH. Then, use the molar ratio from the balanced equation to calculate the amount of H3PO4 needed to react with 25.4g of NaOH.
2 moles of NaOH will react with 1 mole of H2SO4 based on the balanced chemical equation: 2NaOH + H2SO4 -> Na2SO4 + 2H2O.
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
One mole to react with ONE mole CH3COOH (ethaancarbonic acid) because it has only ONE proton (H+) to react with, the other 3 Hydrogen's are not proteolytic (=NON-acidic)
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.
The balanced chemical equation for the reaction is: 3 H3PO4 + 2 NaOH -> Na3PO4 + 3 H2O From the equation, we can see that 3 moles of H3PO4 react with 2 moles of NaOH. Calculate the molar mass of H3PO4 and NaOH. Then, use the molar ratio from the balanced equation to calculate the amount of H3PO4 needed to react with 25.4g of NaOH.
2 moles of NaOH will react with 1 mole of H2SO4 based on the balanced chemical equation: 2NaOH + H2SO4 -> Na2SO4 + 2H2O.
The answer is 50 mL.
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
1337
Since NaOH and HBr react in a 1:1 ratio, the moles of NaOH needed to titrate HBr can be calculated. Moles of NaOH = moles of HBr. Next, use the concentration and volume of HBr to find the moles present. Finally, use the concentration of NaOH to calculate the volume needed. In this case, approximately 2.41 mL of 0.305M NaOH would be needed.
You need to know 3 out of 4 variables to calculate the unknown. Formula: V1*M1=V2*M2 In your question only V1 (=50 mL) and M2 (=0.24m) are known. Eg. V2 (volume of added NaOH in mL) has to be determined (most probably your buret readings; I hope you still can find it back in your lab journal)
The balanced chemical equation for the reaction between HCl and NaOH is: HCl + NaOH -> NaCl + H2O Since the stoichiometry of the reaction is 1:1 for NaCl and HCl, if 1.4 moles of HCl react, then 1.4 moles of NaCl will be formed.
The mass of lead(II) nitrate required to react with 370 g NaOH is 1 531,9 g.
The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) is 1 mol NaOH : 1 mol HCl. Calculate the amount of NaOH required by converting the given mass of HCl to moles using its molar mass, then use the mole ratio from the balanced equation to determine the moles of NaOH needed, and finally convert moles of NaOH to grams using its molar mass.