approx 2.4354g
39.27
0.02 moles of beryllium iodide is equal to 5,256 g.
For this you need the atomic (molecular) mass of KCl. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. KCl= 74.6 grams50.0 grams KCl / (74.6 grams) = .670 moles KCl
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 ===========================
how many moles are there in 56.0 grams of silver nitrate?
530,3 g potassium iodide are needed.
The weight will approximately be 950 grams. YOUR WELCOME.
6.08
39.27
Molar mass of Magnesium Iodide=151.2g/mole 1 Molar solution=151.2g/L 0.5 M solution=75.6g/L=75.6g/1000mL=37.8g/500mL
Find moles potassium iodide first.2.41 grams KI (1 mole KI/166 grams) = 0.01452 moles KIMolarity = moles of solute/Liters of solution ( 100 ml = 0.1 Liters )Molarity = 0.01452 moles KI/0.1 Liters= 0.145 M KI solution================
0.02 moles of beryllium iodide is equal to 5,256 g.
31.6grams
0.02 moles of beryllium diiodide = 5,256 grams
The nitrogen iodide is NI3.
Balanced equation. Zn + I2 --> ZnI2 All is one to one, so it does not matter what is limiting and what drives this reaction. We will use zinc as the driver of convenience. 1 mole Zn (1 mole ZnI2/1 mole Zn)(319.21 grams/1 mole ZnI2) = 319.21 grams zinc iodide produced ==========================
For this you need the atomic (molecular) mass of KCl. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. KCl= 74.6 grams50.0 grams KCl / (74.6 grams) = .670 moles KCl