Answer:
2Fe + 3H2SO4 + Fe2(SO4)3+ 3H2
Explanation:
1. Fe (SO4) 3 is an incorrectly written formula because iron is trivalent as we can see by this three ahead of SO4. SO4 is divalent always.
2. since (SO4) is 3, this three shows us that there must be 3 in the reactants as well.
so now there is 3H2SO4
3. Since we have added 3 to one hydrogen we must add another. So now it's 3H2
4. and finally iron. In Fe2 (SO4) 3 we see this 2 in front of Fe which means it goes 2Fe.
Answer:
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
Explanation:
Step 1: Data given
Nitric acid = HNO3
Molar mass of H = 1.01 g/mol
Molar mass of N = 14.0 g/mol
Molar mass O = 16.0 g/mol
Number of moles nitric acid (HNO3) = 0.25 moles
Molairty = 0.10 M
Step 2: Calculate molar mass of nitric acid
Molar mass HNO3 = Molar mass H + molar mass N + molar mass (3*O)
Molar mass HNO3 = 1.01 + 14.0 + 3*16.0
Molar mass HNO3 = 63.01 g/mol
Step 3: Calculate mass of solute use
Mass HNO3 = moles HNO3 * molar mass HNO3
Mass HNO3 = 0.25 moles * 63.01 g/mol
Mass HNO3 = 15.75 grams
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
<span>The difference between an element and a compound is that an element is a substance made of same type of atoms, whereas a compound is made of different elements in definite proportions. Examples of elements include iron, copper, hydrogen and oxygen. Examples of compounds include water (H2O) and salt (Sodium Chloride - NaCl) </span>
Answer:
308 g
Explanation:
Data given:
mass of Fluorine (F₂) = 225 g
amount of N₂F₄ = ?
Solution:
First we look to the reaction in which Fluorine react with Nitrogen and make N₂F₄
Reaction:
2F₂ + N₂ -----------> N₂F₄
Now look at the reaction for mole ratio
2F₂ + N₂ -----------> N₂F₄
2 mole 1 mole
So it is 2:1 mole ratio of Fluorine to N₂F₄
As we Know
molar mass of F₂ = 2(19) = 38 g/mol
molar mass of N₂F₄ = 2(14) + 4(19) =
molar mass of N₂F₄ = 28 + 76 =104 g/mol
Now convert moles to gram
2F₂ + N₂ -----------> N₂F₄
2 mole (38 g/mol) 1 mole (104 g/mol)
76 g 104 g
So,
we come to know that 76 g of fluorine gives 104 g of N₂F₄ then how many grams of N₂F₄ will be produce by 225 grams of fluorine.
Apply unity formula
76 g of F₂ ≅ 104 g of N₂F₄
225 g of F₂ ≅ X of N₂F₄
Do cross multiplication
X of N₂F₄ = 104 g x 225 g / 76 g
X of N₂F₄ = 308 g
So,
308 g N₂F₄ can be produced from 225 g F₂
