B2Br6+HNO3--->B(NO3)3+HBr

Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide. 4HCl(aq)+MnO2(s)⟶M

zmey [24]

Answer:

HCl is the limiting reactant.

Explanation:

The reactant that yields the lesser number of moles of product is the limiting reactant

Number of moles of MnO2= mass of MnO2/ molar mass of MnO2

Number of moles= 37.7g/86.9368 g/mol= 0.4336 moles

From the reaction equation,

If 1 mole of MnO2 yields 1 mole of chlorine gas

0.4336 moles of MnO2 also yield 0.4336 moles of chlorine gas

For HCl

Number of moles of HCl= mass of HCl/ molar mass of HCl

Number of moles= 46.3g/36.46 g/mol = 1.27 moles of HCl

If 4 moles of HCl yields 1 mole of chlorine gas

1.27 moles of HCl will yield 1.27×1/4 = 0.3175 moles of HCl

Hence HCl is the limiting reactant.

8 0

3 years ago

A 1.00-L gas sample at 100.°C and 500. torr contains 52.0% helium and 48.0% xenon by mass. What are the partial pressures of the

Nataly_w [17]

Answer:

Partial pressure of He=486 torr, partial pressure of Xe= 14 torr

Explanation:

Using the equation, PV=nRT----------------------------------------(1)

Making n the subject of the formula;

n= PV/RT---------------------(2)

Where n= number of moles, v= volume, T= temperature, P= volume.

n= (500 torr/760 torr × 1 atm)× 1L ÷ 373K ×0.082 L atmK^-1. Mol^-1

n= 0.6579 atm.L/ 30.6233

n= 0.0215 mol.

Let the total mass of the gas= 2b(in g).

Mass of helium gas = b (in g) = mass of Xenon gas

Mole of helium gas= b(in g) / 4 gmol^-1

=b/4 mol

Mole of Xenon= b g/131.3 gmol^-1

= b/131.3 mol.

Solving for b, we have;

b/4+b/131.3 = 0.0215 mole

(131.3+4)b/525.2= 0.0215

Multiply both sides by 1/135.3.

b= 11.2918/135.3

b= 0.0835 g

Mole of He gas= 0.0835/4= 0.0209

Mole of Xe gas= 0.0215- 0.0209

= 0.0006 mol

Mole fraction of He = 0.0209/0.0215

= 0.972

Mole fraction of Xe= 0.0006/0.0215

= 0.028

Partial pressure of He gas= 0.972× 500 torr= 486 torr

Partial pressure of Xe gas= 0.028 ×500 torr = 14 torr

4 0

4 years ago

Please help me with my homework..

Snowcat [4.5K]

Answer:

can we talk on my private meeting with you

3 0

3 years ago

Nicotine, a component of tobacco, is composed of C, H, and N. A 5.250-mg sample of nicotine was combusted, producing 14.242 mg o

aev [14]

n - number of mole

M(CO2) = 12.0 +16.0*2 = 44.0 mg/mmol

M(H2O) = 2*1.0 +16.0= 18.0 mg/mmol

1) 14.242 mg of CO2*1 mmol/44.0g =0.32368 mmol CO2

0.32368 mmol CO2 has 0.32368 mmol C

2) 0.32368 mmol C*12 mg/1 mmol= 3.88418 mg C

3) 4.083 mg of H2O * 1 mmol/18 mg = 0.226833 mmol H2O

0.226833 mmol H2O has 2*0.226833 mmol H = 0.453666 mmol H

4) 0.453666 mmol H*1 mg/1mmol = 0.453666 mg H

5) Mass of N = 5.250 mg - 3.88418 mg (C) - 0.453666 mg (H)≈ 0.912 mg N

0.912 mg N * 1 mmol/14.0 mg = 0.0652 mmol N

6) n(C) : n(H) : n(N) = 0.32368 mmol C : 0.453666 mmol H : 0.0652 mmol N

0.0652 mmol N is a smallest value.

n(C) : n(H) : n(N) = (0.32368 mmol/0.0652 mmol) C : (0.453666 mmol H/0.0652 mmol) : (0.0652mmol/0.0652 mmol)N = 5 : 7 : 1

n(C) : n(H) : n(N) = 5 : 7 : 1

Empirical formula : C5H7N.

8 0

3 years ago

How many molecules are in 0.500 mol of AlBr3

mart [117]

.500 mol ( 6.02x10^23 molecules/1 mol) = 3.01x10^23

4 0

3 years ago