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3 years ago

2C4 H10 + 13O2 -→ 8CO2 + 10H2O If 356 moles of O2 are available to react, how many moles of CO2 will fom?

Chemistry

1 answer:

nlexa [21]3 years ago

5 0

Answer:

Answer:

The mole ratio of C₄H₁₀ and CO₂ is 2 : 8, which simplifies to 1 : 4.

Explanation:

The mole ratio is the relative proportion of the moles of products or reactants that participate in the reaction according to the chemical equation.

The chemical equation given is:

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Once you check that the equation is balanced, you can set the mole ratios for all the reactants and products. The coefficients used in front of each reactant and product, in the balanced chemical equation, tells the mole ratios.

In this case, they are: 2 mol C₄H₁₀ : 13 mol O₂ : 8 mol CO₂ : 10 mol H₂O

Since you are asked about the mole ratio of C₄H₁₀ and CO₂ it is:

2 mol C₄H₁₀ : 8 mol CO₂ , which dividing by 2, simplifies to

1 mol C₄H₁₀ : 4 mol CO₂, or

1 : 2.

Explanation:

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The atomic number and the mass number of 18 and 40. Write the information conveyed by this statement.

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There are 18 protons and electrons and 22 neutrons in the atom

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Silicon carbide (SiC) is an important ceramic material made by reacting sand (silicon dioxide, SiO2) with powdered carbon at a h

makkiz [27]

Answer:

Percent yield of SiC is 77.0%.

Explanation:

Balanced reaction: $SiO_{2}+3C\rightarrow SiC+2CO$

Molar mass of SiC = 40.11 g/mol

Molar mass of $SiO_{2}$ = 60.08 g/mol

So, 100.0 kg of $SiO_{2}$ = $\frac{100.0\times 10^{3}}{60.08}$ moles of $SiO_{2}$ = 1664 moles of $SiO_{2}$

According to balanced equation, 1 mol of $SiO_{2}$ produces 1 mol of SiC

Therefore, 1664 moles of $SiO_{2}$ produce 1664 moles of SiC

Mass of 1664 moles of SiC = $(1664\times 40.11)g$ = 66743g = 66.74 kg (4 sig. fig.)

Percent yield of SiC = [(actual yield of SiC)/(theoretical yield of SiC)] $\times 100$ %

= $\frac{51.4kg}{66.74kg}\times 100$ %

= 77.0%

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False. The water would be toxic without purifying it before drinking.

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Express the concentration of a 0.0390 M aqueous solution of fluoride, F − , in mass percentage and in parts per million (ppm). A

Vadim26 [7]

Answer:

Mass percentage → 0.074 %

[F⁻] = 741 ppm

Explanation:

Aqueous solution of flouride → [F⁻] = 0.0390 M

It means that in 1L of solution, we have 0.0390 moles of F⁻

We need the mass of solution and the mass of 0.0390 moles of F⁻

Mass of solution can be determined by density:

1g/mL = Mass of solution / 1000 mL

Note: 1L = 1000mL

Mass of solution: 1000 g

Moles of F⁻ → 0.0390 moles . 19g /1 mol = 0.741 g

Mass percentage → (Mass of solute / Mass of solution) . 100

(0.741 g / 1000 g) . 100 = 0.074 %

Ppm = mass of solute . 10⁶ / mass of solution (mg/kg)

0.741 g . 1000 mg/1g = 741 mg

1000 g . 1 kg/1000 g = 1kg

741 mg/1kg = 741 ppm

5 0

3 years ago