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2 years ago

The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of water.

Chemistry

1 answer:

avanturin [10]2 years ago

6 0

a. 0.137

b. 0.0274

c. 1.5892 g

d. 0.1781

e. 5.6992 g

<h3>Further explanation</h3>

Given

Reaction

2 C4H10 + 13O2 -------> 8CO2 + 10H2O

2.46 g of water

Required

moles and mass

Solution

a. moles of water :

2.46 g : 18 g/mol = 0.137

b. moles of butane :

= 2/10 x mol water

= 2/10 x 0.137

= 0.0274

c. mass of butane :

= 0.0274 x 58 g/mol

= 1.5892 g

d. moles of oxygen :

= 13/2 x mol butane

= 13/2 x 0.0274

= 0.1781

e. mass of oxygen :

= 0.1781 x 32 g/mol

= 5.6992 g

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Answer:

four million two hundred and thirty

Explanation:

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Which correctly lists the types of matter, in terms of their atoms, in order from least tightly packed to most tightly packed? s

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Answer:

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Explanation:

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3 years ago

How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 1.38 g of fe2s3 if the percent yield

ahrayia [7]

Answer: 100 ml


Explanation:

1) Convert 1.38 g of Fe₂S₃ into number of moles, n

i) Formula: n = mass in grass / molar mass

ii) molar mass of Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol


iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of Fe₂S₃


2) Use the percent yield to calculate the theoretical amount:


65% = 0.65 = actual yield/ theoretical yield =>

theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol Fe₂S₃

3) Chemical equation:


 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)

4) Stoichiometrical mole ratios:


3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl

5) Proportionality:

2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃

=> x = 0.020 mol FeCl₃

6) convert 0.020 mol to volume


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3 years ago

Read 2 more answers

How many grams of K2O will be produced from 0.50 g of K and 0.10 g of O2?

Rudik [331]

Answer:

0.6g

Explanation:

Given parameters:

Mass of K = 0.5g

Mass of O₂ = 0.10g

Unknown:

Mass of K₂O = ?

Solution:

To solve this problem, let us write the reaction equation first;

4K + O₂ → 2K₂O

The reaction above delineates the balanced chemical reaction.

To solve this problem, we need to know the limiting reactant. This reactant is the one that determines the amount and extent of the reaction because it is given in short supply. The other reactant is the one in excess.

Start off by find the number of moles of the reactant;

Number of moles = $\frac{mass}{molar mass}$

Molar mas of K = 39g/mol

Molar mass of O₂ = 2(16) = 32g/mol

Number of moles of K = $\frac{0.5}{39}$ = 0.013moles

Number of moles of O₂ = $\frac{0.1}{32}$ = 0.031moles

From the balanced reaction;

4 moles of K reacted with 1 mole of O₂

0.013 moles of K will react with $\frac{0.013}{4}$ = 0.0078 moles of O₂

We see that oxygen gas is in excess. We were given 0.031moles of the gas but only require 0.0078moles of oxygen gas.

The limiting reactant is potassium.

therefore;

4 moles of K produced 2 moles of K₂O

0.013 moles of K will produce $\frac{0.013 x 2}{4}$ = 0.0065‬moles of K₂O

to find the mass of K₂O;

Mass of K₂O = number of moles x molar mass

Molar mass of K₂O = 2(39) + 16 = 94g/mol

Mass of K₂O = 0.0065 x 94 = 0.6g

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