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2 years ago

Calculate the mass of Octane needed to release 6.20 mol Co2

1 answer:

n200080 [17]2 years ago

8 0

The combustion reaction of octane is as follow,

C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O

According to balance equation,

8 moles of CO₂ are released when = 114.23 g (1 mole) Octane is reacted

So,

6.20 moles of CO₂ will release when = X g of Octane is reacted

Solving for X,
X = (114.23 g × 6.20 mol) ÷ 8 mol

X = 88.52 g of Octane
Result:
88.52 g of Octane is needed to release 6.20 mol CO₂.

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What is the composition of tungsten?

Setler79 [48]

Answer:

90% to 97% pure tungsten in a matrix of nickel and copper or nickel and iron.

Explanation:

Heavy metal tungsten alloys are 90% to 97% pure tungsten in a matrix of nickel and copper or nickel and iron. The addition of these alloying elements improves both the ductility and machinability of these alloys over non-alloyed tungsten.

7 0

2 years ago

Read 2 more answers

Please help, I really don’t understand this!!!

kap26 [50]

<u>Analysing the Question:</u>

We are given the balanced equation:

C₆H₁₂O₆ + 6O₂→ 6CO₂ + 6H₂O

from this equation, we can say that: <em>for every 1 mole of Glucose, we need 6 moles of Oxygen</em>

<u>Moles of Glucose used in the reaction:</u>

Molar mass of Glucose = 180 grams / mol

Given mass of Glucose = 1 gram

Mole of Glucose = Given mass / Molar mass

Moles of Glucose = 1 / 180 moles

<u>Mass of Oxygen required:</u>

We know that for every mole of Glucose, we need 6 moles of Oxygen

So, for 1/180 moles of Glucose, we need 6 / 180 = 1 / 30 moles of Oxygen

Mass of 1 / 30 moles of Oxygen:

Mass = Molar mass * number of moles

Mass of Oxygen = 32 * 1/30

Mass of Oxygen = 32 / 30

Mass of Oxygen = 1.06 grams

5 0

3 years ago

Type the correct answer in the box. Express your answer to three significant figures. Calcium nitrate reacts with sodium phospha

RoseWind [281]

6.07 grams is the theoretical yield of calcium phosphate (Ca₃(PO₄)₂).

<h3>How we calculate mass from moles?</h3>

Mass of any substance can be calculated by using moles as:

n = W/M, where

W = required mass

M= molar mass

Given chemical reaction is:

3Ca(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + Ca₃(PO₄)₂

From the stoichiometry it is clear that:

3 moles of Ca(NO₃)₂ = produce 1 mole of Ca₃(PO₄)₂

Given mass of Ca(NO₃)₂ = 96.1g

Mole of Ca(NO₃)₂ = 96.1g/164g/mol = 0.5859moles

So, 0.5859 moles of Ca(NO₃)₂ = produce 0.5859×1/3 = 0.0196 moles of Ca₃(PO₄)₂

Required mass of Ca₃(PO₄)₂ will be calculated by using moles as:

W = 0.0196mole × 310g/mole = 6.07 grams

Hence, 6.07 grams is the theoretical yield of calcium phosphate.

To know more about moles, visit the below link:

brainly.com/question/15373263

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2 years ago

Answer question 13. What is the mechanical advantage of a lever that can lift a 100 N load with a input force of 20 N

vazorg [7]

MA= output force/ input force
MA= 100N/20N
MA= 50

8 0

3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago