All categories

2 years ago

How many moles of H2O can be produced with 5.8 moles of O2 ?

Chemistry

1 answer:

Evgesh-ka [11]2 years ago

8 0

Answer:

1.78

Explanation:

1mole produce 18g/x will produce 32g then the answer is 1.78moles

You might be interested in

What is the mass of 531 ml of a liquid that has a density of 0.760 g/ml?

pogonyaev

Mass is density times volume
0.760 x 531 = 403.56g
404g if 3 sig figs

4 0

3 years ago

When an acid or vase is added to to water,

velikii [3]

Answer:

yes

Explanation:

3 0

3 years ago

What thype of bond requires the give and take of electrons ?

DedPeter [7]

Ionic bond involves electrostatic attraction between oppositely charged ions.
The ions are atoms that have gained 1 or more electrons and atoms that have lost 1 or more electrons.
<span>Answer: The type of bond that requires the give and take of electrons is Ionic bond</span>

6 0

3 years ago

Read 2 more answers

What can be said of a closed system when an exothermic reaction proceeds in an aqueous solution?

lyudmila [28]

An exothermic reaction is a type of reaction that dissipates heat as the reaction proceeds. This would mean that in a closed system, when a reaction proceeds and is endothermic, the temperature of the solution or the system would increase so as to maintain the equilibrium with the whole system.

8 0

3 years ago

I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r

maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0

3 years ago