using the equation 2h2+o2-->2h2o, if 192g of oxygen are produced, how many grams of hydrogen must react with it

1 answer:

4 0

The balanced chemical reaction is:

<span>2H2+O2-->2H2O
</span>
To determine the mass of hydrogen that is needed, we need use the initial amount of oxygen and relate it to hydrogen from the reaction given. We do as follows:

192 g O2 ( 1 mol O2 / 32 g O2) ( 2 mol H2 / 1 mol O2 ) ( 2.02 g H2 / 1 mol H2 ) = 24.24 g H2

You might be interested in

The price of tomatoes is 1.8 lb/$. How many $ would you need to buy 8.1 lb?

sesenic [268]

If you divide 8.1 by 1.8 that leads to 4.5

7 0

2 years ago

In 1986 an electrical power plant in taylorsville, georgia, burned 8,376,726 tons of coal, a national record at that time. assum

RideAnS [48]

6 0

3 years ago

Read 2 more answers

PLEASE HELP!!!!!! WILL GIVE BRAINLIEST!!!! IT'S SUPER EASY

Nana76 [90]

the primary consumer in those photos would be C

6 0

2 years ago

Felix adds salt to the boiling water, what will he have in the pot?

Pani-rosa [81]

A mixture called a solution, I think.

8 0

2 years ago

An iron(iii) sulfate hydrate is 18.4% water. What is the formula of the hydrate? What is the name of the hydrate?

aleksley [76]

Answer:- Formula of the hydrate is $Fe_2(SO_4)_3.5H_2O$ and it's name is Iron(III)sulfate pentahydrate.

Solution:- As per the given information, there is 18.4% water in the hydrate. If we assume the mass of the hydrate as 100 grams then there would be 18.4 grams of water and 81.6 grams of Iron(III)sulfate present in the hydrate.

Molar mass for Iron(III)sulfate is 399.88 gram per mol and the molar mass for water is 18.02 gram per mol.

We will calculate the moles of Iron(III)sulfate and water present in the compound on dividing their grams by their molar masses as:

$81.6gFe_2(SO_4)_3(\frac{1mol}{399.88g})$

= $0.204molFe_2(SO_4)_3$

$18.4gH_2O(\frac{1mol}{18.02g})$

= $1.02molH_2O$

Now, the next step is to calculate the mol ratio and for this we divide the moles of each by the least one of them means whose moles are less. Here, the moles of Iron(III)sulfate are less than moles of water. So, we divide the moles of each by 0.204.

$Fe_2(SO_4)_3=\frac{0.204}{0.204}$ = 1

$H_2O=\frac{1.02}{0.204}$ = 5

There is 1:5 mol ratio between Iron(III)sulfate and water. So, the formula of the hydrate is $Fe_2(SO_4)_3.5H_2O$ and the name of the hydrate is Iron(III)sulfate pentahydrate.

3 0

2 years ago