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klemol [59]
3 years ago
9

Carbon disulfide is prepared industrially by reacting carbon with sulfur dioxide according to the above equation. If 70.8 g of c

arbon react, how many moles of CS2 are produced? Show all work, include all units, and box your answer.
Chemistry
1 answer:
Rus_ich [418]3 years ago
8 0

Answer:

1.18 moles of CS₂ are produced by the reaction.

Explanation:

We present the reaction:

5C + 2SO₂  →  CS₂  +  4CO

5 moles of carbon react to 2 moles of sulfur dioxide in order to produce 1 mol of carbon disulfide and 4 moles of carbon monoxide.

As we do not have data from the SO₂, we assume this as the excess reagent. We convert the mass of carbon to moles:

70.8 g / 12 g/mol = 5.9 moles

Ratio is 5:1, so 5 moles of carbon react to produce 1 mol of CS₂

Then, 5.9 moles will produce (5.9 . 1) / 5 = 1.18 moles

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If you have 12g of compound A2B3 (159.7g/mol), how many moles do you have?
ololo11 [35]

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0.075 moles n=m/M so divide the mass (m) by the molar mass (M) to get the n which is the number of moles

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3 years ago
Part 1. Determine the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm. Show your work.
zaharov [31]

If a sample of gas is a 0.622-gram, volume of 2.4 L at 287 K and 0.850 atm. Then the molar mass of the gas is 7.18 g/mol

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases.

An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given :

  • V = 2.4 L = 0.0024
  • P = 86126.25 Pa
  • T =  287 K
  • m = 0.622
  • R = 8.314

The ideal gas equation is given below.

n = PV/RT

n = 86126.25 x 0.0024 / 8.314 x 287

n = 0.622 / molar mass (n = Avogardos number)

Molar mass =  7.18 g

Hence, the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm is 7.18 g

More about the ideal gas equation link is given below.

brainly.com/question/4147359

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4 0
2 years ago
Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta
Brut [27]

Answer:

The molar mass of unknown gas is 145.82 g/mol.

Explanation:

Volume of oxygen gas effused under time t = 8.24 mL

Effusion rate of oxygen gas = R=\frac{8.24 mL}{t}

Molar mass of oxygen gas = 32 g/mol

Volume of unknown gas effused under time t = 3.86 mL

Effusion rate of unknown gas = R'=\frac{3.86 mL}{t}

Molar mass of unknown gas = M

Graham's Law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

\frac{R}{R'}=\sqrt{\frac{M}{32 g/mol}}

\frac{\frac{8.24 mL}{t}}{\frac{3.86 mL}{t}}=\sqrt{\frac{M}{32 g/mol}}

M=\frac{32 g/mol\times 8.24 \times 8.24}{3.86\times 3.86}=145.82 g/mol

4 0
3 years ago
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