Which of these equations is balanced? (Select all that are correct)

Calculate the number of moles in 7g of nitrogen gas

Andru [333]

Answer:

Explanation:

7gN2 x $\frac{1 mol}{2(14.01) g}$

=0.2498mol N2

Nitrogen gas has the formula $N_{2}$ so therefore that means you would have to multiply the mass in the molar by 2. To solve for the number of moles you need to cancel out the grams, you do this by using the molar mass of nitrgoen gas. You get the value on in the denominator from the periodic table (atomic mass of element). The grams will cancel out, leaving you with the number of moles when you divide 7/2(14.01).

5 0

3 years ago

When octane (C8H18) is burned in the presence of oxygen, the yield of products (carbon dioxide and water) is 87%. What mass of c

ahrayia [7]

Answer:

14.5g of CO₂ are produced

Explanation:

The reaction of octane with oxygen is:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O

<em>Where 1 mole of octane (Molar mass: 114.23g/mol) reacts with 25/2 moles of O₂ (Molar mass 32g/mol) to produce 8 moles of CO₂ and 9 moles of water.</em>

When 21.0 g of octane is burned with 19.0 g of oxygen gas you need to find <em>limiting reactant </em>to find how many moles of products are formed:

Octane: 21.0g ₓ (1mol / 114.23g) = 0.184 moles octane

Oxygen: 19.0g ₓ (1 mol / 32g) = 0.594 moles oxygen

For a complete reaction of 0.184 moles of octane you will need:

0.184 moles C₈H₁₈ ₓ (25/2 moles O₂ / 1 mole C₈H₁₈) = <em>2.3 moles of oxygen</em>

As you have just 0.594 moles of oxygen, <em>Oxygen is limiting reactant.</em>

Based on chemical equation, 25/2 of O₂ produce 8 moles of CO₂, that means theoretical yield of CO₂ with 0.594 moles of O₂ is:

0.594 moles O₂ ₓ (8 moles CO₂ / 25/2 moles O₂) = 0.380 moles of CO₂

But, as yield of products is 87%, moles produced of CO₂ are:

0.380 moles of CO₂ ₓ 87% = 0.331 moles CO₂ are produced.

As molar mass of CO₂ is 44g/mol, mass of CO₂ in 0.331 moles is:

0.331 moles CO₂ ₓ (44g / mol) =

<h3>14.5g of CO₂ are produced</h3>

6 0

3 years ago

How many moles of hydrogen are in 5.2 moles of c7h18

alexandr1967 [171]

The percentage of hydrogen in C7H18 is calculated as follows:
[18/(12*7+1*8)]*100=18%
The amount of hydrogen in 5.2moles is given by:(18/100)*5.2=0.94moles

6 0

3 years ago

Read 2 more answers

Also need to know some of this stuff

irinina [24]

Answer:

answers are in attachment

Explanation:

1. Synthesis Reaction: In this type of reaction multiple reactant combine to form a single product.

2. Decomposition Reaction: In this type of reaction single compound or reactant break down into new elements or compounds.

3. Replacement Reaction: In this type of reaction an element replace an element in a compound.

4. Even hydrogen is not a metal but it can act as metal in single replacement reaction.

5. In combustion reaction fuel react with oxygen and give heat and light that increase temperature of surrounding.

6. In in-complete combustion reaction fuel react with in-sufficient oxygen and give carbon monoxide and carbon in form of soot.

7. Base is a compound that liberate OH⁻ ion in water.

8. indicators have different color in acid and base depends on pH of the solution

9. Salt and water are the product of neutralization reaction

NaOH + HCl = NaCl + H₂O

10. The pH of neutral solution is 7.

6 0

3 years ago

What is the total volume of gaseous products formed when 116 liters of butane (C4H10) react completely according to the followin

Contact [7]

<u>Answer:</u> The total volume of the gaseous products is 1044.29 L

<u>Explanation:</u>

We are given:

Volume of butane = 116 L

At STP:

22.4 L of volume is occupied by 1 mole of a gas

So, 116 L of volume will be occupied by = $\frac{1}{22.4}\times 116=5.18mol$ of butane

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

<u>For carbon dioxide:</u>

By Stoichiometry of the reaction:

2 moles of butane produces 8 moles of carbon dioxide

So, 5.18 moles of butane will produce = $\frac{8}{2}\times 5.18=20.72mol$ of carbon dioxide

Volume of carbon dioxide at STP = (20.72 × 22.4) = 464.13 L

<u>For water vapor:</u>

By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water vapor

So, 5.18 moles of butane will produce = $\frac{10}{2}\times 5.18=25.9mol$ of water vapor

Volume of water vapor at STP = (25.9 × 22.4) = 580.16 L

Total volume of the gaseous products = [464.13 + 580.16] = 1044.29 L

Hence, the total volume of the gaseous products is 1044.29 L

3 0

3 years ago