700 L of water was produced if 350.0 L of carbon dioxide were made at STP.
The quantitative relationship (ratio) between reactants and products in a chemical reaction that produces gases is known as gas stoichiometry. When the created gases are presumed to be ideal and their temperature, pressure, and volume are all known, gas stoichiometry is applicable.
The ideal gas equation is PV=nRT, where n is the number of moles and R is the gas constant, P is the pressure measured in atmospheres (atm), V is the volume measured in liters (L), and
Calculations based on stoichiometry assist scientists and engineers who work in the business world in estimating the number of items they will make using a particular process. They can also assist in determining if a product will be economical to produce.
Reduced growth, reproduction, and survivability for the consumer are typically the results of a significant stoichiometric imbalance between the primary producer and consumer.
To know more about stoichiometry refer to: brainly.com/question/9743981
#SPJ1
Solution :
From the balanced chemical equation, we can say that 1 moles of KBr will produce 1 moles of KCl .
Moles of KBr in 102 g of potassium bromide.
n = 102/119.002
n = 0.86 mole.
So, number of miles of KCl produced are also 0.86 mole.
Mass of KCl produced :
Hence, this is the required solution.
Ok what you want. I got u
Answer : The partial pressure of 
and 
are, 84 torr and 778 torr respectively.
Explanation : Given,
Mass of = 15.0 g
Mass of = 22.6 g
Molar mass of = 197.4 g/mole
Molar mass of = 32 g/mole
First we have to calculate the moles of and .
and,
Now we have to calculate the mole fraction of and .
and,
Now we have to partial pressure of and .
According to the Raoult's law,
where,
= partial pressure of gas
= total pressure of gas
= mole fraction of gas
and,
Therefore, the partial pressure of and are, 84 torr and 778 torr respectively.
Answer:
-3.7771 × 10² kJ/mol
Explanation:
Let's consider the following equation.
3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)
where,
n: moles
ΔG°f(): standard Gibbs free energy of formation
p: products
r: reactants
ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))
ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)
ΔG° = -377.71 kJ = -3.7771 × 10² kJ
This is the standard Gibbs free energy per mole of reaction.
