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3 years ago

Please Help Me!!!

Chemistry

1 answer:

sergejj [24]3 years ago

7 0

Answer:

The products are carbon dioxide and water

Explanation:

Step 1: Data given

Combustion = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve O2 as one reactant.

Step 2: The complete combustion of C3H7OH:

For the combustion of 1-propanol, we need O2.

The products of this combustion are CO2 and H2O.

C3H7OH + O2→ CO2 + H2O

On the left side we have 3x C (in c3H7OH), on the right side we have 1x C (in CO2). To balance the amount of C, we have to multiply CO2 on the right side by 3

C3H7OH + O2→ 3CO2 + H2O

On the left side we have 8x H (in C3H7OH) and 2x on the right side (in H2O). To balance the amount of H, we have to multiply H2O, on the right side by 4.

C3H7OH + O2→ 3CO2 + 4H2O

On the left side we have 3x O (1x in C3H7OH and 2x in O2), on the right side we have 10x O (6x in CO2 and 4x in H2O).

To balance the amount of O on both sides, we have to multiply C3H7OH by 2, multiply O2 by 9. Then we have to multiply 3CO2 by 2 and 4H2O by 2. Now the equation is balanced.

2C3H7OH + 9O2→ 6CO2 + 8H2O

For 2 moles propanol, we need 9 moles of O2 to produce 6 moles of CO2 and 8 moles Of H2O

The products are carbon dioxide and water

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Copper(II) sulfide is formed when copper and sulfur are heated together. In this reaction,

storchak [24]

Answer:

The mass of copper(II) sulfide formed is:

= 81.24 g

Explanation:

The Balanced chemical equation for this reaction is :

$Cu(s) + S\rightarrow CuS$

given mass= 54 g

Molar mass of Cu = 63.55 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$moles=\frac{54}{63.55}$

Moles of Cu = 0.8497 mol

Given mass = 42 g

Molar mass of S = 32.06 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$moles=\frac{42}{32.06}$

Moles of S = 1.31 mol

Limiting Reagent : The reagent which is present in less amount and consumed in a reaction

First find the limiting reagent :

$Cu + S\rightarrow CuS$

1 mol of Cu require = 1 mol of S

0.8497 mol of Cu should require = 1 x 0.8497 mol

= 0.8497 mol of S

S present in the reaction Medium = 1.31 mol

S Required = 0.8497 mol

S is present in excess and Cu is limiting reagent

All Cu is consumed in the reaction

Amount Cu will decide the amount of CuS formed

$Cu + S\rightarrow CuS$

1 mole of Cu gives = 1 mole of Copper sulfide

0.8497 mol of Cu = 1 x 0.8497 mole of Copper sulfide

= 0.8497

Molar mass of CuS = 95.611 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$0.8497 = \frac{given\ mass}{95.611}$

Mass of CuS = 0.8497 x 95.611

= 81.24 g

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