If you start with 5 molecules of O2 in the reaction 2Mg+O2--> 2MgO, how much Mg will you need?

earnstyle [38]

Answer:

Alcohols are usually named by the first procedure and are designated by an -ol suffix, as in ethanol, CH3CH2OH (note that a locator number is unnecessary on a two-carbon chain). On longer chains the location of the hydroxyl group determines chain numbering. For example: (CH3)2C=CHCH(OH)CH3 is 4-methyl-3-penten-2-ol.10

Explanation:

3 0

2 years ago

Read 2 more answers

275 g of a compound with the molecular formula ccl2f2 contains how many molecules of the compound

madreJ [45]

brainly.com/question/8581600#respond

3 0

3 years ago

I need this answer quick please show work

Ainat [17]

Answer:

The answer to your question is 25.2 g of acetic acid.

Explanation:

Data

[Acetic acid] = 0.839 M

Volume = 0.5 L

Molecular weight = 60.05 g/mol

Process

1.- Calculate the number of moles of acetic acid

Molarity = moles / volume

-Solve for moles

moles = Molarity x volume

-Substitution

moles = (0.839)(0.5)

-Result

moles = 0.4195

2.- Calculate the mass of acetic acid using proportions and cross multiplications

60.05 g ----------------------- 1 mol

x ----------------------- 0.4195 moles

x = (0.4195 x 60.05) / 1

x = 25.19 g

3.- Conclusion

25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M

4 0

3 years ago

Limiting Reactants—————-

denis-greek [22]

Answer:

21.8 grams.

Explanation:

Molar mass data from a modern periodic table:

Mg: 24.301;
O: 15.999.

How many moles of MgO will be produced if Mg is the limiting reactant?

Number of moles of Mg:

$\displaystyle n = \frac{m}{M} = \frac{16.3}{24.301} = 0.670644\;\text{mol}$ .

The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.

How many moles of MgO will be produced if O₂ is the limiting reactant?

Number of moles of O₂:

$\displaystyle n = \frac{m}{M} = \frac{4.33}{15.999} = 0.270642\;\text{mol}$ .

The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. $2\times 0.270642 = 0.541284\;\text{mol}$ of MgO will be produced if O₂ is in excess.