1+7+(8x3)
=32
=32x2 (2 moles)
=64
The molar mass of nitric acid (HNO3) is 63.01 g/mol. To find the total grams in 4 moles, you would multiply the molar mass by the number of moles: 63.01 g/mol x 4 mol = 252.04 grams. So, there would be 252.04 grams in four moles of nitric acid (HNO3).
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.
Starting with the formula: 2HNO3 --> H2O + NO2 If you have 0.4 moles of nitric acid (HNO3), you will get half the number of moles of NO2. So, you will have 0.2 moles of nitric acid.
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.
Starting with the formula: 2HNO3 --> H2O + NO2 If you have 0.4 moles of nitric acid (HNO3), you will end up with half the moles of nitrogen dioxide (NO2)...so you will have 0.2 moles.
The molar mass of nitric acid (HNO3) is 63.01 g/mol. To find the total grams in 4 moles, you would multiply the molar mass by the number of moles: 63.01 g/mol x 4 mol = 252.04 grams. So, there would be 252.04 grams in four moles of nitric acid (HNO3).
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.
Starting with the formula: 2HNO3 --> H2O + NO2 If you have 0.4 moles of nitric acid (HNO3), you will get half the number of moles of NO2. So, you will have 0.2 moles of nitric acid.
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.
Starting with the formula: 2HNO3 --> H2O + NO2 If you have 0.4 moles of nitric acid (HNO3), you will end up with half the moles of nitrogen dioxide (NO2)...so you will have 0.2 moles.
The molar mass of nitric acid (HNO3) is approximately 63 grams per mole.
To determine the number of moles of calcium hydroxide needed to react with the nitric acid, you would need to know the concentration of the nitric acid. With the concentration, you can use the balanced chemical equation of the reaction to calculate the moles of calcium hydroxide required.
The balanced chemical equation for the reaction between potassium hydroxide (KOH) and nitric acid (HNO3) is 1:1 ratio. Therefore, 3 moles of nitric acid will require 3 moles of potassium hydroxide to neutralize it.
To neutralize the nitric acid, you need a 1:1 mole ratio of sodium hydroxide to nitric acid. First, calculate the moles of nitric acid in the solution using the formula Molarity = moles/volume. Then, use the mole ratio to find the moles of sodium hydroxide needed. Finally, convert this to grams using the molar mass of sodium hydroxide.
Mg2+(s) + 2HNO3(l)= Mg(NO3)2(aq) + H2(g) since the only mole value given is 8 I must assume this is the limiting reactant. Because of the 2:1 ratio of Nitric acid to Magnesium Nitrate, meaning there must be 2 moles Nitric acid for every 1 mole Magnesium Nitrate formed, 4 moles of Magnesium nitrate will be formed.
The mass of sulfuric acid is 490,395 grams.
I think you meant " How many moles of acetic acid in 25 grams of acetic acid? " We will use the chemist formula for acetic acid, 25 grams C2H4O2 (1 mole C2H4O2/60.052 grams) = 0.42 mole acetic acid =================