Answer:
2.2 moles of Fe will be produced
Explanation:
Step 1: Data given
Number of moles of hydrogen gas = 3.3 moles
Number of moles of iron oxide = 1.5 moles
Step 2: The balanced equation
3H2 + Fe2O3 → 2Fe + 3H2O
Step 3: Calculate the limiting reactant
For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O
Hydrogen gas is the limiting reactant. It will completely be consumed (3.3 moles). Fe2O3 is in excess. There will react 3.3 / 3 = 1.1 moles
There will remain 1.5 - 1.1 = 0.4 moles Fe2O3
Step 4: Calculate moles Fe
For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O
For 3.3 moles H2 we'll have 2/3 * 3.3 = 2.2 moles Fe
2.2 moles of Fe will be produced
Amplitude is the awnser glad I could help you
Answer:
the l sign means the substance in the chemical equation is a liquid. (hope this helped : D )
Missing question: 0,535 gram of KIO₃ dissolved in 250 mL of de-ionized water to <span>make primary standard solution.
m(</span>KIO₃) = 0,535 g.
V(KIO₃) = 250 mL ÷ 1000 mL/L = 0,25 L.
n(KIO₃) = m(KIO₃) ÷ M(KIO₃).
n(KIO₃) = 0,535 g ÷ 214 g/mol.
n(KIO₃) = 0,0025 mol.
c(KIO₃) = n(KIO₃) ÷ V(KIO₃).
c(KIO₃) = 0,0025 mol ÷ 0,25 L.
c(KIO₃) = 0,01 mol/L = 0,01 M.
Answer:
- The answer is the concentration of an NaOH = 1.6 M
Explanation:
The most common way to solve this kind of problem is to use the formula
In your problem,
For NaOH
C₁ =?? v₁= 78.0 mL = 0.078 L
For H₂SO₄
C₁ =1.25 M v₁= 50.0 mL = 0.05 L
but you must note that for the reaction of NaOH with H₂SO₄
2 mol of NaOH raect with 1 mol H₂SO₄
So, by applying in above formula
- (C₁ * 0.078 L) = (2* 1.25 M * 0.05 L)
- C₁ = (2* 1.25 M * 0.05 L) / (0.078 L) = 1.6 M
<u>So, the answer is the concentration of an NaOH = 1.6 M</u>
