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irina1246 [14]
2 years ago
15

How many molecules of O₂ will be required to produce 28.8 g of water? 2H₂ + O₂ ---> 2H₂O

Chemistry
1 answer:
juin [17]2 years ago
4 0

To produce 28.8 g of water 4.81 x 10²³ molecules of O₂ will be required

<h3>What is limiting reagent ?</h3>

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

According to given equation;

2H₂ + O₂ ---> 2H₂O

If 36 g of H₂O (2 mole) is produced by 1 mole O₂ i.e, 32 gram.

Therefore, 28.8 g of H₂O is produced by ;

32/36 x 28.8 = 25.6 g of O₂

Number of molecules of O₂ = given weight / molecular weight x 6.023 x 10²³

                                              = 25.6 / 32 x 6.023 x 10²³

                                              = 4.81 x 10²³ molecules of O₂

Therefore,4.81 x 10²³ molecules of O₂ will be required to produce 28.8 g of water.

Learn more about Limiting reagent here ;

brainly.com/question/20070272

#SPJ1

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how many moles of hydrogen gas will be produced if you start with 2.5 moles of magnesium and an excess of hydrochloric acid give
Vinil7 [7]

Number of moles of Hydrogen gas produced = 2.5 moles

Given the balanced chemical equation,

                   Mg+2HCl \rightarrow MgCl_2+H_2

This means, 1 mole of Mg and 2 moles of HCl gives 1 mole of Hydrogen gas.

Given that 2.5 moles of Magnesium is taken and an excess of Hydrochloric acid.

The mole ratio of Magnesium and Hydrogen = 1:1

Thus 2.5 moles of Magnesium with an excess of Hydrochloric acid will yield 2.5 moles of Hydrogen gas.

[By referring excess of HCl, we may assume that it contains at least 5 moles of HCl. Then only 2.5 moles of Mg will produce 2.5 moles of Hydrogen gas].

Learn more about the mole ratio at brainly.com/question/19099163

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6 0
2 years ago
A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution
Marianna [84]

Answer:

0.136g

Explanation:

A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?

Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)

Initial mole of Co(NO3)2  =\frac{mass}{molar mass}

=\frac{5.00}{182.94} \\\\=0.02733mol

Mole of Co(NO3)2 in final solution

=\frac{4.00}{100}\times 0.02733\\\\=0.04\times 0.02733\\\\= 0.001093mol

Mole of  NO3- in final solution = 2 x Mole of Co(NO3)2

=2\times 0.001093\\\\=0.002186mol

Mass of  NO3- in final solution is mole x Molar mass of NO3

=0.002186\times62.01\\\\=0.136g

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3 years ago
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Answer:  6100 seconds

Explanation:                                                                                                            6.1 Kiloseconds = 6100 Seconds

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