We calculate for the number of moles of water given its mass by dividing the given mass by the molar mass.
n water = (36.04 g) / (18 g/mol)
n water = 2 mols
From the given balanced equation, every 6 moles of water produced will require 7 moles of oxygen.
n oxygen = (2 mols H2O) x (7 moles O2 / 6 moles H2O)
n oxygen = 2.33 mols O2
Answer:
-1815.4 kJ/mol
Explanation:
Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:
∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)
This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.
In this case:
ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol
Answer:
7.46 g
Explanation:
From the balanced equation, 2 moles of Mg is required for 2 moles of MgO.
The mole ratio is 1:1
mole = mass/molar mass
mole of 4.50 g Mg = 4.50/24.3 = 0.185 mole
0.185 mole Mg will tiled 0.185 MgO
Hence, theoretical yield of MgO in g
mass = mole x molar mass
0.185 x 40.3 = 7.46 g
Answer:
0.467 kilograms of protein or carbohydrates
Explanation:
First, there is a need to understand that <em>1 kJ = 0.239 kCal</em>
Hence, 17 kJ = 17 x 0.239 = 4.063 kCal
The race requires 650.0 kCal/hr and has to last for 175 minutes.
175 minutes = 175/60 = 2.917 hrs
The total kCal requires for the race = 650 x 2.917 = 1,895.833 kCal
1 g of protein or carbohydrate food produces 17.0 kJ or 4.063 kCal of energy. Hence, the total g of protein or carbohydrate that will produce 1,895.833 kCal of energy would be;
1,895.833/4.063 = 466.609 g
<em>1 g = 0.001 kg</em>
466.609 g = 466.609 x 0.001 = 0.467 kg
<em>Hence, </em><em>0.467 </em><em>kilograms of proteins or carbohydrates must be consumed.</em>
