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3 years ago

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What volume of "laughing gas" (N2O) will be produced from 60 g of nitrogen gas (N2) and 85 g of oxygen gas (O2)? How much of the

excess reactant will be left? 2N2 + O2 -> 2N2O

1 answer:

riadik2000 [5.3K]3 years ago

3 0

Answer:

94.16g of N2O

50.88g of O2

Explanation:

O2 reacts with N2 as follows:

2N2 + O2 → 2N2O

To find moles of N2O produced we need to convert the mass of each gas to moles to find limiting reactant and the moles of N2O that could be produced:

<em>Moles N2 (Molar mass: 28g/mol):</em>

60g N2 * (1mol / 28g) = 2.14 moles N2

<em>Moles O2 (Molar mass: 32g/mol):</em>

85g O2 * (1mol / 32g) = 2.66 moles O2

For a complete reaction of 2.14 moles of N2 are needed:

2.14 mol N2 * (1mol O2 / 2 mol N2) = 1.07 moles of O2.

As there are 2.66 moles of O2, reactant in excess is O2 and will remain:

2.66 moles - 1.07 moles =

1.59 moles O2. The mass is:

1.59 moles O2 * (32g / mol) =

<h3>50.88g of O2</h3><h3 />

And there are produced:

2.14 mol N2 * (2 moles N2O / 2 moles N2) = 2.14 moles N2O.

The mass is (44g/mol for N2O):

2.14 moles N2O * (44g/mol) =

<h3>94.16g of N2O</h3>

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Determine the freezing point of a water solution of fructose (C6H12O6) made by dissolving 92.0 g of fructose in 202 g of water.

Naya [18.7K]

Answer:

THE FREEZING POINT OF A WATER SOLUTION OF FRUCTOSE MADE BY DISSOLVING 92 g OF FRUCTOSE IN 202g OF WATER IS -4.70 ◦C

Explanation:

To calculate the freezing point of a water solution of fructose,

1. calculate the molar mass of Fructose

( 12 * 6 + 1*12 + 16*6) =72 + 12 +96 = 180 g/mol

2. calculate the number of moles of fructose in the solution

number of moles = mass / molar mass

n = 92 g / 180 g/mol

n = 0.511 moles.

3. calculate the molarity of the solution

molarity = moles / mass of water in kg

molarity = 0.5111 / 202 g /1000 g

molarity = 0.5111 / 0.202

molarity = 2.529 M

4. calculate the change in the freezing point of pure solvent and solution ΔTf

ΔTf = Kf * molarity of the solute

Kf = 1.86 ◦C/m for water

ΔTf = 1.86 * 2.529

ΔTf = 4.70 C

5. the freezing point is therefore

0.00 ◦C - 4.70 ◦C = -4.70 ◦C

4 0

3 years ago

Read 2 more answers

What is the final temperature when 150.0 g of water at 90.0 °c is added to 100.0 g of water at 30.0 OC? Note that C, of water ca

levacccp [35]

Answer : The final temperature is, $337.8K$

Explanation :

$Q_{absorbed}=Q_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of water at $90^oC$ = 150 g

$m_2$ = mass of water at $30^oC$ = 100 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of lead = $90^oC=273+90=363K$

$T_2$ = temperature of water = $30^oC=273+30=303K$

$c_1\text{ and }c_2$ = same (for water)

Now put all the given values in equation (1), we get

$150\times (T_{final}-363)=-[100\times (T_{final}-303)]$

$T_{final}=337.8K$

Therefore, the final temperature is, $337.8K$

8 0

4 years ago

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Use molecular orbital theory to determine whether He2 2+ or He2 + is more stable. Use molecular orbital theory to determine whet

julia-pushkina [17]

Answer:

The He₂ 2+ ion is more stable since it has a higher bond order (bond order = 1) than the He₂ + ion (bond order = 1/2).

Explanation:

Molecular orbital of He₂⁺

$1\sigma_{1s}^21\sigma(star)_{1s}^1$

There are two electrons in bonding and 1 electron in antibonding orbital

Bond order = $\frac{(2-1)}{2}$

= $\frac{1}{2}$

Molecular orbital of He₂⁺²

$1\sigma_{1s}^21\sigma(star)_{1s}^0$

There are two electrons in bonding and 0 electron in antibonding orbital

Bond order = $\frac{(2-0)}{2}$

= 1

So bond order of He₂⁺² is 1 which is more stable than He₂⁺ whose bond order is $\frac{1}{2}$ .

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3 years ago

Copper(II) sulfide, CuS, is used in the development of aniline black dye in textile printing. What is the maximum mass of CuS wh

IrinaK [193]

Explanation:

The chemical reaction equation will be as follows.

$CuCl_{2} + (NH_{4})_{2}S \rightarrow 2NH_{4}Cl + CuS$

As 1 mole $CuCl_{2}$ reacts with 1 mole of $(NH_{4})_{2}S$

Hence, moles of $CuCl_{2}$ = Molarity × Volume (in mL)

= 0.5 × 38.0 mL

= 0.019 mol

Moles of $(NH_{4})_{2}S$ = Molarity × Volume (in mL)

= 0.6 × 42.0 mL

= 0.0252 mol

As moles of CuS is less than the moles of $(NH_{4})_{2}S$ . This means that CuS is the limiting reagent.

Thus, maximum moles of CuS formed are 0.019 mol. As molar mass of CuS is 95.6 g/mol.

Therefore, maximum mass of CuS can be formed is $95.6 g/mol \times 0.019 mol$

= 1.8164 g

Thus, we can conclude that maximum mass of CuS formed is 1.8164 g.

6 0

3 years ago

Write the formulas of the following compounds:

dem82 [27]

Answer:

a) Li2CO3

b) NaCLO4

c) Ba(OH)2

d) (NH4)2CO3

e) H2SO4

f) Ca(CH3COO)2

g) Mg3(PO4)2

f) Na2SO3

Explanation:

a) 2Li + CO3 ↔ Li2CO3

b) NaOH * HCLO4 ↔ NaCLO4 + H2O

c) Ba + 2H2O ↔ Ba(OH)2 +

d) 2NH4 + H2CO3 ↔ (NH4)2CO3 + H2O

c) SO2 + NO2 +H2O ↔ H2SO4 + NOx

f) 2CH3COOH + CaO ↔ Ca(CH3COOH)2 + H2O

g) 3MgO + 2H3PO4 ↔ Mg3(PO4)2 + H2O

h) NaOH + H2SO3 ↔ Na2SO3 + H2O

6 0

3 years ago