All categories

2 years ago

What is the percent by mass of oxygen?

Chemistry

1 answer:

Colt1911 [192]2 years ago

3 0

Answer:

For oxygen: mass % O = (mass of 1 mol of oxygen/mass of 1 mol of CO2) x 100. mass % O = (32.00 g / 44.01 g) x 100. mass % O = 72.71 %

Explanation:

You might be interested in

The energy exchange with the environment during a chemical reaction is called the heat of reaction.

JulijaS [17]

This is a hard question

6 0

3 years ago

How many grams of potassium nitrate (KNO3) would form if 2.25 liters of a 1.50 molar lead nitrate Pb(NO3)2 solution reacts with

fgiga [73]

I know that the answer is 639 g

4 0

3 years ago

Name four products of incomplete combustion

aalyn [17]

Answer: The products are carbon monoxide, water and carbon.

Explanation:

5 0

2 years ago

This is agree or disagree I need help ASAP!!

julsineya [31]

Answer:

disagree

sometimes they attract

Explanation:

8 0

2 years ago

Read 2 more answers

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

kirill115 [55]

Hello!

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

We have the following data:

m(Fe) - mass of iron = 28 g

m(S) - mass of sufur = 16 g

MM(Fe) - molar mass of iron ≈ 56 g/mol

MM(S) - molar mass of sulfur ≈ 32 g/mol

n(Fe) - number of mol of iron = ?

n(S) - number of mol of sulfur = ?

Solving:

* to n(Fe)

$n_{Fe} = \dfrac{m_{Fe}}{MM_{Fe}}$

$n_{Fe} = \dfrac{28\:\diagup\!\!\!\!\!g}{56\:\diagup\!\!\!\!\!g/mol}$

$\boxed{n_{Fe} = 0.5\:mol}$

* to n(S)

$n_{S} = \dfrac{m_{S}}{MM_{S}}$

$n_{S} = \dfrac{16\:\diagup\!\!\!\!\!g}{32\:\diagup\!\!\!\!\!g/mol}$

$\boxed{n_{S} = 0.5\:mol}$

The stoichiometric reaction will be in the same proportion (1 : 1), let us see:

$Fe + S \Longrightarrow FeS$

1 mol of Fe -------------- 1 mol of FeS

0.5 mol of Fe ------------ 0.5 mol of FeS

Will the reaction of the iron mass with the mass of sulfur produce how many grams of iron sulfide? We will see:

n(FeS) - number of mol of iron sulfide = 0.5 mol

m(FeS) - mass of iron sulfide = ? (in grams)

MM(FeS) - Molar Mass of iron sulfide = 56 + 32 = 88 g/mol

Solving:

$n_{FeS} = \dfrac{m_{FeS}}{MM_{FeS}}$

$m_{FeS} = n_{FeS}*MM_{FeS}$

$m_{FeS} = 0.5\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}*88\:\dfrac{g}{mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}}$

$\boxed{\boxed{m_{FeS} = 44\:g}}\:\:\:\:\:\:\bf\blue{\checkmark}\bf\green{\checkmark}\bf\red{\checkmark}$

Answer:

44 grams of iron sulfide

___________________________

$\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

4 0

2 years ago