Consider the chemical reaction: 2 H2 O(l) ⟶2 H2(g)+O2(g) What mass of H2 O is required to form 1

Consider the chemical reaction: 2 H2 O(l) ⟶2 H2(g)+O2(g)...

Consider the chemical reaction: $$ 2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ What mass of $\mathrm{H}_{2} \mathrm{O}$ is required to form $1.4 \mathrm{~L}$ of $\mathrm{O}_{2}$ at a temperature of $315 \mathrm{~K}$ and a pressure of $0.957 \mathrm{~atm}$ ?

Consider the chemical reaction:
$$
2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)
$$
What mass of $\mathrm{H}_{2} \mathrm{O}$ is required to form $1.4 \mathrm{~L}$ of $\mathrm{O}_{2}$ at a temperature of $315 \mathrm{~K}$ and a pressure of $0.957 \mathrm{~atm}$ ?

Key Concepts

Ideal Gas Law

The Ideal Gas Law, expressed as PV = nRT, relates the pressure, volume, temperature, and moles of a gas. This law is essential for converting the measured gas volume at given temperature and pressure conditions to the number of moles, a necessary step in many gas-related calculations.

Molar Mass

Molar mass is the mass of one mole of a substance and serves as the conversion factor between the number of moles and the mass of a substance. It is critical for translating the mole-based results obtained from stoichiometric calculations into practical mass quantities.

Stoichiometry

Stoichiometry involves using a balanced chemical equation to determine the quantitative relationships between reactants and products. It allows the calculation of how much of one substance is needed to react completely with another, based on their molar ratios derived from the equation.

Transcript

00:01 Alright guys and i'm back again question about 72 i'm using the reaction given to calculate the mass of water that is required to form 1 .4 liter of oxygen at 315 caffeine and 0 .957 atmosphere so i have the reaction to liquid giving me 2 hydrogen gas and plus oxygen.

00:40 So 1 .4 volume of oxygen form equals to 1 .4 later at temperature of 315 kelvin and pressure of 0 .957 atmosphere.

01:09 I want to calculate the mass of water that is required to form this volume of oxygen.

01:19 At this temperature so my approach to softening this question is using the gas law general gas law so i want to calculate the volume of this oxygen at standard temperature and pressure we are given the volume of this oxygen at this temperature you know standard temperature temperature is 298 kivin and pressure will be one atmosphere so i want to calculate the equivalent of this volume at this temperature from there i can use the molar volume to that give me i divert the mole of this the mole of oxygen that is from once i know the volume at this temperature at standard condition i can get the more of the more of oxygen that is formed once i know the volume at this temperature at of oxygen i can get the mold of water and i can get the mass of water so i will make i will call this one my v2 this one will be my v1 this one will be my v1 this one will be my t1 this one will be my p1 and this one will be my t2 and p2 so i want to calculate my v2 so v2 will be p1 v1 t2 b5 by b2 t1 second substitute that introduce the pressure and that is 0 .957 atmosphere but by by v1 is 1 .4 liter and then t1 is 3 oh t2 is 298 kelvin 27 to 98 kavine 35 by p2 which is 1 atmosphere multiplied by t1 which is 3 15 and that gives me b2 it costs 1 .16 116 later so the oxygen that is for we have volume of 1 .16 later at stp so i will use that to calculate so i will use the molar volume of the cup without to calculate the mass of water form so i will set it up like this 1 .16 later of oxygen at stp we have one more of oxygen we have 22 .4 later volume at its oxygen one more of oxygen oxygen at stp we have volume of 22 .4 liter so this will give me the more the mole of 1 .16 liter of oxygen then i also have the relationship from the reaction how one more of oxygen one more of oxygen from this reaction one more of oxygen is formed from two more of water and then we know one more of water as a mass of 18 gram.

06:02 So this is water is h2...

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