The answer is going to be 476.06.
This compound is Boron selenate. Molar mass of B2(SeO4)3 is 450.4948 g/mol.
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
1) Answer is: c) The reaction will proceed right.
Balanced chemical reaction: N₂(g) + 3H₂(g) ⇄ 2NH₃(g) ΔH = +92 kJ.
Reducing the volume of the system increase the partial pressures of the products and reactants.
With a pressure increase due to a decrease in volume, the side of the equilibrium with fewer moles is more favorable, there are 4 moles at the left side (three moles of hydrogen and one mole of nitrogen) and 2 moles (ammonia) at the right side of the reaction.
2) Answer is: d) The partial pressure of ammonia will increase.
This reaction is endothermic (enthalpy is higher than zero), which means that heat is added.
According to Le Chatelier's principle when the reaction is endothermic heat is included as a reactant and when the temperature increased, the heat of the system increase, so the system consume some of that heat by shifting the equilibrium to the right, producing more ammonia.
