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3 years ago

C3H8+5O2 →3CO2+4H2

Chemistry

1 answer:

Lisa [10]3 years ago

3 0

Answer:

The answer to your question is 9.075 g of CO₂

Explanation:

Data

mass of C₃H₈ = 39 g

mass of O₂ = 11 g

Balanced chemical reaction

C₃H₈ + 5O₂ ⇒ 3CO₂ + 4H₂

-Calculate the molar mass of the reactants

C₃H₈ = (12 x 3) + (8 x 1) = 36 + 8 = 44 g

O₂ = (16 x 2) = 32 g

-Calculate the limiting reactant

theoretical yield C₃H₈ / O₂ = 44/5(32) = 44/ 160 = 0.275

experimental yield C₃H₈/O₂ = 39/11 = 3.5

From the previous result, we conclude that the limiting reactant is O₂ because the experimental yield was higher than the theoretical yield.

-Calculate the mass of CO₂

160 g of O₂ ----------------- 3(44) g of CO₂

11 g of O₂ ------------------ x

x = (11 x 3(44)) / 160

x = 1452 / 160

x = 9.075 g of CO₂

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3 years ago

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Consider the following unbalanced chemical equation. C5H12(l) + O2(g) → CO2(g) + H2O(l) If 21.9 grams of pentane (C5H12) are bur

Marina CMI [18]

Answer : The mass of water produced will be 32.78 grams.

Explanation : Given,

Mass of $C_5H_{12}$ = 21.9 g

Molar mass of $C_5H_{12}$ = 72.15 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of $C_5H_{12}$ .

$\text{Moles of }C_5H_{12}=\frac{\text{Mass of }C_5H_{12}}{\text{Molar mass of }C_5H_{12}}=\frac{21.9g}{72.15g/mole}=0.3035moles$

Now we have to calculate the moles of $H_2O$ .

The balanced chemical reaction will be,

$C_5H_{12}(l)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(l)$

From the balanced reaction we conclude that

As, 1 mole of $C_5H_{12}$ react to give 6 moles of $H_2O$

So, 0.3035 moles of $C_5H_{12}$ react to give $0.3035\times 6=1.821$ moles of $H_2O$

Now we have to calculate the mass of $H_2O$ .

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=(1.821mole)\times (18g/mole)=32.78g$

Therefore, the mass of water produced will be 32.78 grams.

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2 years ago

How far can you travel in 7 hours if you are driving at a rate of 55 mph

Oksi-84 [34.3K]

the answer is 385 miles

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2 years ago

Determine whether you can swim in 1.00 x 10^27 molecules of water.

zloy xaker [14]

Answer:

We can not swim in 1.00 × 10²⁷ molecules of water

Explanation:

The given number of molecules of water = 1.00 × 10²⁷ molecules

The Avogadro's number, $N_A$ , gives the number of molecules in one mole of a substance

$N_A$ ≈ 6.0221409 × 10²³ molecules/mol

Therefore

Therefore, we have;

The number of moles of water present in 1.00 × 10²⁷ molecules, n = (The number of molecules of water) ÷ $N_A$

∴ n = (1.00 × 10²⁷ molecules)/(6.0221409 × 10²³ molecules/mol) = 1,660.53902857 moles

The mass of one mole of water = The molar mass of water = 18.01528 g/mol

The mass, 'm', of water in 1,660.53902857 moles of water is given as follows;

Mass = (The number of moles of the substance) × (The molar mass of the substance)

∴ The mass of the water in the given quantity of water, m = 1,660.53902857 moles × 18.01528 g/mol ≈ 29.9150756 kg.

The density pf water, ρ = 997 kg/m³

Volume = Mass/Density

∴ The volume of the water present in the given quantity of water, v = 29.9150756 kg/(997 kg/m³) ≈ 30.0050909 liters

The volume of the water present in 1.00 × 10²⁷ molecules of water ≈ 30.0 liters

The average volume of a human body = 62 liters

Therefore, we can not swim in the given quantity of 1.00 × 10²⁷ molecules = 30.0 liters water

7 0

2 years ago

A 25.0-mL sample of 0.150 M butanoic acid is titrated with a 0.150 M NaOH solution. What is the pH before any base is added? The

hoa [83]

HA ⇄ H⁺ + A⁻
so:
$\frac{[H^+][A^-]}{[HA]} = 1.5 x 10^{-5}$
and now:
$\frac{(x)(x)}{(0.150-x)}$ = 1.5 x 10⁻⁵
x is considered very small compared to 0.15
x² = 2.25 x 10⁻⁶
x = 1.5 x 10⁻³
So [H⁺] = 1.5 x 10⁻³
pH = - log [H⁺] = - log (1.5 x 10⁻³) = 2.83

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3 years ago