All categories

4 years ago

If you burn 32.3g of hydrogen and produce 288g of water, how much oxygen reacted?

Chemistry

1 answer:

viva [34]4 years ago

7 0

Answer:

256 g

Step-by-step explanation:

a) Balanced equation

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 2.016 32.00 18.02

2H₂ + O₂ ⟶ 2H₂O

m/g: 32.3 288

(b) Moles of H₂O

n = 288 g H₂O × (1 mol H₂O/18.02 g H₂O)

= 15.98 mol H₂O

(c) Moles of O₂

The molar ratio is (2 mol H₂O /1 mol O₂)

n = 15.98 mol H₂O × (1 mol O₂ /2 mol H₂O)

= 7.991 mol O₂

(d) Mass of O₂

m = 7.991 mol O₂× (32.00 g O₂/1 mol O₂)

= 256 g O₂

The reaction requires 256 g O₂.

Note: If you base your answer on the mass of hydrogen, you get the same result.

Mass of O₂ = 32.3 g H₂ × 1 mol H₂/2.016 g H₂

= 16.02 mol H₂ × 1 mol O₂ /2 mol H₂

= 8.011 mol O₂ × 32.00 g O₂/1 mol O₂

= 256 g O₂

You might be interested in

A sample weighing 3.110 g is a mixture of Fe 2 O 3 (molar mass = 159.69 g/mol) and Al 2 O 3 (molar mass = 101.96 g/mol). When he

grandymaker [24]

Answer:

The mass fraction of ferric oxide in the original sample : $\frac{723}{3110}$

Explanation:

Mass of the mixture = 3.110 g

Mass of $Fe_2O_3=x$

Mass of $Al_2O_3=y$

After heating the mixture it allowed to react with hydrogen gas in which all the ferric oxide reacted to form metallic iron and water vapors where as aluminum oxide did not react.

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

Mass of mixture left after all the ferric oxide has reacted = 2.387 g

Mass of mixture left after all the ferric oxide has reacted = y

$x=3.110 g- y=3.110 g - 2.387 g = 0.723 g$

The mass fraction of ferric oxide in the original sample :

$\frac{0.723 g}{3.110 g}=\frac{723}{3110}$

5 0

3 years ago

What is chemical reaction?

Aliun [14]

Answer:

a process that involves rearrangement of the molecular or ionic structure of a substance, as distinct from a change in physical form or a nuclear reaction

3 0

3 years ago

Read 2 more answers

Given the balanced equation, 2 H2 + O2 -> 2 H2O, answer the following

9966 [12]

Answer: $O_2$ is the limiting reagent

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} H_2=\frac{10.0g}{2g/mol}=5.00moles$

$\text{Moles of} O_2=\frac{20.0g}{32g/mol}=0.625moles$

The balanced chemical reaction is :

$2H_2+O_2\rightarrow 2H_2O$

According to stoichiometry :

1 moles of $O_2$ require = 2 moles of $H_2$

Thus 0.625 moles of $O_2$ will require= $\frac{2}{1}\times 0.625=1.25moles$ of $H_2$

Thus $O_2$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent as it is present more than the required amount.

8 0

3 years ago

An objects mechanical energy is the sum of its?

frosja888 [35]

Mechanical energy of an object can be the result of its motion
i think

3 0

4 years ago

Draw the product formed by the reaction of potassium t-butoxide with (1s,2s)-1-bromo-2-methyl-1-phenylbutane (shown below). (dra

VikaD [51]

You have to use a Newman projection to make sure that the H on C#2 is anti-coplanar with the Br on C#1. (Those are the two things that are going to be eliminated to make the alkene.)

My Newman projection looks like this when it's in the right configuration:
Front carbon (C#2) has ethyl group straight up, H down/left, and CH3 down/right
Back carbon (C#1) has H straight down, Ph up/left, and Br up/right.

Then when you eliminate the H from C#2 and the Br from C#1, you will have Ph and the ethyl group on the same side of the molecule, and you'll have the remaining H and CH3 on the same side of the molecule.

This is going to give you (Z)-2-methyl-1-phenyl-1-butene.

3 0

3 years ago

Read 2 more answers

If you burn 32.3g of hydrogen and produce 288g of water, how much oxygen reacted?​

If you burn 32.3g of hydrogen and produce 288g of water, how much oxygen reacted?