To determine the number of moles in 4g of sodium hydroxide (NaOH), divide the mass by the molar mass of NaOH. The molar mass of NaOH is approximately 40g/mol (Na = 23g/mol, O = 16g/mol, H = 1g/mol). Therefore, 4g of NaOH is equal to 0.1 moles.
60 g NaOH x 1 mole NaOH/40 g NaOH = 1.5 moles NaOH
To determine the number of moles in 20g of sodium hydroxide, you need to divide the given mass by the molar mass of sodium hydroxide. The molar mass of NaOH is 40 g/mol (sodium: 23 g/mol, oxygen: 16 g/mol, hydrogen: 1 g/mol). So, 20g NaOH / 40 g/mol = 0.5 moles of sodium hydroxide.
First, calculate the number of moles of nitric acid present in 3.50 L of 0.700 M solution. Since nitric acid is a diprotic acid, the mole ratio with sodium hydroxide is 1:2. Then, use the mole ratio to determine the number of moles of sodium hydroxide needed to neutralize the nitric acid. Finally, convert the moles of sodium hydroxide to grams using its molar mass.
To solve this stoichiometry problem, first calculate the number of moles of sodium hydroxide (NaOH) present in 200 grams. Then, using the balanced equation, determine the moles of sodium sulfate (Na2SO4) that will be formed. Finally, convert the moles of Na2SO4 to grams using the molar mass of sodium sulfate.
The balanced chemical equation for the neutralization between sodium hydroxide (NaOH) and nitric acid (HNO3) is 1 mol of NaOH reacts with 1 mol of HNO3. Therefore, 20 moles of nitric acid would require 20 moles of sodium hydroxide to neutralize it.
In the acid-base reaction where sodium hydroxide and sulfuric acid react, the formula is: H2SO4 + 2NaOH --> Na2SO4 + 2H2O. The coefficients shown are necessary to uphold the law of conservation of mass. So, if you have 17 moles of sulfuric acid, you will need twice as many moles of sodium hydroxide, so the answer is 34 moles NaOH.
No amount of sodium sulphate can be formed from sodium hydroxide alone, because sodium sulfate contains sulfur and sodium hydroxide does not. By neutralization with sulphuric acid, one formula unit of sodium sulphate can be formed from two moles of sodium hydroxide, according to the equation 2 NaOH + H2SO4 -> Na2SO4 + 2 H2O.
To calculate the grams of sodium hydroxide present in the solution, first calculate the number of moles using the formula: moles = Molarity (M) x Volume (L). Then, use the molar mass of sodium hydroxide (NaOH) to convert moles to grams. The molar mass of NaOH is 40 g/mol. Thus, in this case, you have 0.3375 moles of NaOH and if you convert this to grams, it would be 13.5 grams.
0.2 mol
60 g NaOH x 1 mole NaOH/40 g NaOH = 1.5 moles NaOH
Sodium reacts with water. 0.652 NaOH moles will form.
To determine the number of moles in 20g of sodium hydroxide, you need to divide the given mass by the molar mass of sodium hydroxide. The molar mass of NaOH is 40 g/mol (sodium: 23 g/mol, oxygen: 16 g/mol, hydrogen: 1 g/mol). So, 20g NaOH / 40 g/mol = 0.5 moles of sodium hydroxide.
The molecular weight of sodium hydroxide is 40g/mol. To get the amount of moles, you have to divide the weight by molecular mass. 12g / 40 is 0.3 moles. This is 300 millimoles.
First, calculate the number of moles of nitric acid present in 3.50 L of 0.700 M solution. Since nitric acid is a diprotic acid, the mole ratio with sodium hydroxide is 1:2. Then, use the mole ratio to determine the number of moles of sodium hydroxide needed to neutralize the nitric acid. Finally, convert the moles of sodium hydroxide to grams using its molar mass.
1. Three moles of sodium contain 18,06642387.1023 atoms. 2. The mass of three moles of sodium is 68,97 grams.
To solve this stoichiometry problem, first calculate the number of moles of sodium hydroxide (NaOH) present in 200 grams. Then, using the balanced equation, determine the moles of sodium sulfate (Na2SO4) that will be formed. Finally, convert the moles of Na2SO4 to grams using the molar mass of sodium sulfate.
The balanced chemical equation for the neutralization between sodium hydroxide (NaOH) and nitric acid (HNO3) is 1 mol of NaOH reacts with 1 mol of HNO3. Therefore, 20 moles of nitric acid would require 20 moles of sodium hydroxide to neutralize it.