Answer:
Explanation:
First, find the mass of empirical formula, CH. 12.01 g/mol is for carbon, and 1.008 g/mol is for hydrogen. 12.01+1.008=13.018 G/mol CH. Divide 78.110 G/mol by 13.018 g/mol. You get approximately 6. Multiply that by the subscript of each element. 6(CH)=
Answer:
Total pressure = 4.57 atm
Explanation:
Given data:
Partial pressure of nitrogen = 1.3 atm
Partial pressure of oxygen = 1824 mmHg
Partial pressure of carbon dioxide = 247 torr
Partial pressure of argon = 0.015 atm
Partial pressure of water vapor = 53.69 kpa
Total pressure = ?
Solution:
First of all we convert the units other into atm.
Partial pressure of oxygen = 1824 mmHg / 760 = 2.4 atm
Partial pressure of carbon dioxide = 247 torr / 760 = 0.325 atm
Partial pressure of water vapor = 53.69 kpa / 101 = 0.53 atm
Total pressure = Partial pressure of N + Partial pressure of O + Partial pressure of CO₂ + Partial pressure of Ar + Partial pressure of water vapor
Total pressure = 1.3 atm + 2.4 atm + 0.325 atm + 0.015 atm + 0.53 atm
Total pressure = 4.57 atm
Answer:
grams H₂O produced = 8.7 grams
Explanation:
Given 2C₂H₆(g) + 7O₂(g) => 4CO₂(g) + 6H₂O(l)
7g 18g ?g
Plan => Convert gms to moles => determine Limiting reactant => solve for moles water => convert moles water to grams water
Moles Reactants
moles C₂H₆ = 7g/30g/mol = 0.233mol
moles O₂ = 18g/32g/mol = 0.563mol
Limiting Reactant => (Test for Limiting Reactant) Divide mole value by respective coefficient of balanced equation; the smaller number is the limiting reactant.
moles C₂H₆/2 = 0.233/2 = 0.12
moles O₂/7 = 0.08
<u><em>Limiting Reactant is O₂</em></u>
Moles and Grams of H₂O:
Use Limiting Reactant moles (not division value) to calculate moles of H₂O.
moles H₂O = 6/7(moles O₂) = 6/7(0.562) moles H₂O = 0.482 mole H₂O yield
grams H₂O = (0.482mol)(18g·mol⁻¹) = 8.7 grams H₂O
