All categories

2 years ago

The combustion of propane (C3H8) produces CO₂ and H₂O:

Chemistry

2 answers:

fomenos2 years ago

5 0

Answer:

4.4 moles of water

Explanation:

Propane is C3H8. Is the chemical equation, you wrote C2H8 which is wrong. Type everything properly and give a check before finally posting the question.

The chemical equation that is given, is not properly balanced. The BALANCED equation would be

C3H8 + 5O2 → 3CO2 + 4H2O

From this balanced chemical equation, we see that 5 moles of oxygen produce 4 moles of water.

5 moles of Oxygen → 4 moles of water

1mole of oxygen → $\frac{4}{5}$ moles of water

5.5 moles of oxygen → ( $\frac{4}{5}$ × 5.5) moles of water = 4.4 moles of water

∴ The reaction of 5.5 mol of O2 will produce 4.4 mol of H2O. [Answer:]

Simora [160]2 years ago

3 0

Answer:

5.5

Explanation:

First, you will need to balance the equation ( the equation in your question is not balanced....typo?)

C2 H8 + 4 O2 ===> 2 CO2 + 4 H2O

Now, from the balanced equation:

O2 : H2O = 4/4 = 1/1

SO 5.5 MOLES OF O2 PRODUCES 5.5 MOLES H20

You might be interested in

It takes 38mL of 0.75M NaOH solution to completely neutralize 155 mL of a chloric acid solution (HclO3). what is the concentrati

Scrat [10]

ITS ABOUT DRIVE
ITS ABOUT POWER
WE STAY HUNGRY
WE DEVOUR
PUT IN THE WORK
PUT IN THE HOURS ⌚
AND TAKE WHATS OURS

8 0

3 years ago

2K + 2HBr → 2 KBr + H2

Inessa [10]

Answer:

$\large \boxed{\text{0.0503 g}}$

Explanation:

The limiting reactant is the reactant that gives the smaller amount of product.

Assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r: 39.10 80.41 2.016

2K + 2HBr ⟶ 2KBr + H₂

m/g: 5.5 4.04

a) Limiting reactant

(i) Calculate the moles of each reactant

$\text{Moles of K} = \text{5.5 g} \times \dfrac{\text{1 mol}}{\text{31.10 g}} = \text{0.141 mol K}\\\\\text{Moles of HBr} = \text{4.04 g} \times \dfrac{\text{1 mol}}{\text{80.91 g}} = \text{0.049 93 mol HBr}$

(ii) Calculate the moles of H₂ we can obtain from each reactant.

From K:

The molar ratio of H₂:K is 1:2.

$\text{Moles of H}_{2} = \text{0.141 mol K} \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol K}} = \text{0.0703 mol H}_{2}$

From HBr:

The molar ratio of H₂:HBr is 3:2.

$\text{Moles of H}_{2} = \text{0.049.93 mol HBr } \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol HBr}} = \text{0.024 97 mol H}_{2}$

(iii) Identify the limiting reactant

HBr is the limiting reactant because it gives the smaller amount of NH₃.

b) Excess reactant

The excess reactant is K.

c) Mass of H₂

$\text{Mass of H}_{2} = \text{0.024 97 mol H}_{2} \times \dfrac{\text{2.016 g H}_{2}}{\text{1 mol H}_{2}} = \textbf{0.0503 g H}_{2}\\ \text{The mass of hydrogen is $\large \boxed{\textbf{0.0503 g}}$ }$

3 0

3 years ago

What is the pH of a solution of 0.600 M K2HPO4, potassium hydrogen phosphate?

AleksAgata [21]

When pKas of polyprotic intermediates have a difference of 2 or more you just average them using the equation: pH = (pKa2 + pKa3) / 2 
pKa2 = -log(Ka2) ; pKa3 = -log(Ka3) 
so, for this problem, REGARDLESS OF THE CONCENTRATION GIVEN, the answer is: 
pH = (7.2076+12.3767) / 2 
pH = 9.79

5 0

4 years ago

Read 2 more answers

In the EXPLORE section of your lesson 4.08 on Potential energy there were several animations to watch that provided a graphic il

Lunna [17]

Answer:

This is because no energy is being created or destroyed in this system

Explanation:

I think this is correct? I hope it helps.

7 0

3 years ago

Read 2 more answers