Answer:
The limiting reactant is oxygen gas and the reaction would produce 3.932 x 10 ^ 7 moles of water
Explanation:
Step 1: Convert everything into moles
nH2(l) = 1.06 x 10^8 g / 2.016 g/mol = 5.26 x 10^7 mols
nO2(l) = 6.29 x 10^8 g / 32.00 g/ mol = 1.966 x 10^7 mols
Step 2: Find the limiting reagent
The limiting reagent would be oxygen gas from
the balanced equation because we have less moles of oxygen gas needed to fully combust with the hydrant gas
Step 3: Stoichiometry time
The mole ratio from oxygen gas to water is 1:2
This means that for every mole of oxygen gas two moles of water is produced
We need to multiply the moles of oxygen gas by two to find out how many moles of water has been produced
nH2O = nO2 x 2
nH2O = 1.966x10^7 x 2
nH2O = 3.932x10^7
Step 4: Therefore statement
Therefore the limiting reactant is oxygen gas and the reaction would produce 3.932 x 10 ^ 7 moles of water
Answer:
1. 276 g of NO₂
2. 34.8 moles of LiO
3. 4.23×10²⁵ molecules of SO₂
4. 540 g of H₂O
5. 224 g CO
Explanation:
Let's define the molar mass of the compound to define the moles or the grans of each.
Molar mass . moles = Mass
Mass (g) / Molar mass = Moles
1. 6 mol . 46 g / 1 mol = 276 g of NO₂
2. 800 g . 1mol / 22.94 g = 34.8 moles of LiO
3. To determine the number of molecules, we convert the mass to moles and then, we use the NA (1 mol contains 6.02×10²³ molecules)
4500 g . 1mol / 64.06 g = 70.2 moles of SO₂
70.2 mol . 6.02×10²³ molecules / 1 mol = 4.23×10²⁵ molecules of SO₂
4. 30 mol . 18g / 1 mol = 540 g of H₂O
5. 8 mol . 28g / 1mol = 224 g CO
Maybe b might be the answer
Answer:
In the quantum-mechanical model of an atom, electrons in the same atom that have the same principal quantum number (n) or principal energy level are said to occupy an electron shell of the atom. Orbitals define regions in space where you are likely to find electrons.
