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3 years ago

Number 6 CO2 + H₂O → C6H12O6 + O2

Chemistry

1 answer:

horrorfan [7]3 years ago

3 0

Explanation:

$\pmb{\sf{\purple{ 6CO_{2}+6H_{2}O \longrightarrow C_{6}H_{12}O_{6}+6O_{2}}}}$

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3 years ago

If you run the reaction to make water: 2H2 + O2 → 2H2O, starting with 3.0 moles of hydrogen gas and 2.0 moles of oxygen gas, wha

Alex_Xolod [135]

Answer:

54 g is the theoretical yield

Explanation:

This is the reaction:

2H₂ + O₂ → 2H₂O

So 2 moles of hydrogen react with 1 mol of oxygen, to produce 2 mol of water.

If I have 3 moles of H₂ and 2 moles of O₂, the my limiting reactant is the hydrogen.

1 mol of O₂ react with 2 moles of H₂

S 2 mol of O₂ would react with 4 moles (I only have 3 moles)

Then, ratio is 2:2 the same as 1:1

As 2 mol of H₂ produce 2 moles of water, 3 moles of H₂ will produce 3 moles of H₂O.

This is the theoretical yield in moles. Let's convert them to mass (mol . molar mass)

3 mol . 18g/m = 54 g

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3 years ago

Read 2 more answers

What is the mass of carbon monoxide (CO) gas if the gas occupies a volume of 4.1 L at 2.0 atm of pressure and -73oC?

Stels [109]

Answer:

1.7

Explanation:

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2 years ago

What name is given to the minimum amount of energy needed by particles for them to react when they collide?

makkiz [27]

Answer:

Activation

Explanation:

8 0

3 years ago

Vitamin C contains only carbon, hydrogen, and oxygen. When a 1.00 g was combusted, 1.4991 g of CO2 and 0.4092 g of H2O were obta

Alina [70]

The empirical formula for this vitamin : C₃H₄O₃

<h3>Further explanation </h3>

The empirical formula is the smallest comparison of atoms of compound =mole ratio of the components

The principle of determining empirical formula

Determine the mass ratio of the constituent elements of the compound.
Determine the mole ratio by dividing the percentage by the atomic mass

Mass of C in CO₂ :(MW C = 12 g/mol, CO₂=44 g/mol)

$\tt \dfrac{12}{44}\times 1.4991=0.409~g$

Mass of H in H₂O :(MW H = 1 g/mol, H₂O = 18 g/mol)

$\tt \dfrac{2.1}{18}\times 0.4092=0.0455~g$

Mass O = Mass sample - (mass C + mass H) :

$\tt 1-(0.409+0.0455)=0.5455~g$

mol ratio C : H : O =

$\tt \dfrac{0.409}{12}\div \dfrac{0.0455}{1}\div \dfrac{0.5455}{16}\\\\0.0341\div 0.0455\div 0.0341\rightarrow 1\div 1.33\div 1=3\div 4\div 3$

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3 years ago