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3 years ago

Examine the reactants of the incomplete synthesis reaction. 2Mg + O2 → __________.

2 answers:

4 0

2MgO
This is called redox reaction. Oxygen is the oxidizing agent to receive 2 electrons from Mg.
Mg is the reducing agent to give 2 electrons to O.
Mg becomes Mg2+ and O becomes O2-. These 2 ions form an ionic structure called MgO

Degger [83]3 years ago

3 0

Answer: $2MgO$

Explanation:

Synthesis reaction is a chemical reaction in which two reactants are combining to form one product.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$2Mg+O_2\rightarrow 2MgO$

Thus the products of chemical reaction will be $2MgO$

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Propane (C3H8) can be burned to produce heat for homes. The products of the reaction are CO2 and H2O. For complete combustion to

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Answer:

1- 3 Moles of CO2

2- 132 g of CO2

3- 105,6 g of CO2

4- Limiting Reagent O2

Products form based on limiting reagent (384g O2) :

CO2: 316,8 g

H2O: 172,8 g

Products form based on C3H8 (132,33 g):

CO2: 396,99 g

H2O: 216,54 g

Explanation:

Atomic Masses:

C: 12

H: 1

O: 16

Molecular weights:

C3H8: 44 g

O2: 32 g

H2O: 18 g

CO2: 44 g

C3H8 + 5 O2⇒ 3 CO2 + 4 H2O

C3H8 (44g)+ O2 (160 g) ⇒ CO2 (132 g) + H2O (72 g)

In 5 moles of O2 are produced 3 moles of CO2, equivalent to 132 g

For 160g of O2 are produced 132 g of CO2, so 128 g of O2

160 g O2 ⇒ 132 g CO2

128 g O2⇒ × = 105,6 g CO2

(128×132÷160= 105,6)

The limiting reagent is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, because the reaction cannot continue without it.

If I have 132,33 g of C3H8 and 384 g O2 we can calculate:

For 44 g of CH3H8 ⇒160 g of O2

With 132,33 g of CH3H8 ⇒ ×= 481,2 g of O2

(132,33×160÷44=481,2)

As this amount exceeds the quantity of O2 that we have, we can assume that the 384 g O2 will be totally consumed.

Calculations of the products formed in base of quantity of O2 (limiting reagent):

160 g O2 ⇒ 132 g CO2

384 g O2⇒ × = 316,8 g CO2

(384×132÷160= 316,8)

160 g O2 ⇒ 72 g H2O

384 g O2⇒ × = 172,8 g H2O

(384×72÷160= 172,8)