Please help me thanks

How many moles would be equal to 2.6 liters of calcium nitrite

lorasvet [3.4K]

6 moles to equal that I do t know for sure though

7 0

3 years ago

How many moles of propane gas would be present in 11 grams of the gas at standard conditions?

Akimi4 [234]

The molar mass is usually referred to with
M
, while the mass is referred to as
m
. The amount of substance is
n
. This gives you the following relationship:
=
M
=
m
n

Since you have given (C3H8)=11 g
m
(
C
3
H
8
)
=
11

g
and you already looked up (C3H8)=44.1 gmol−1
M
(
C
3
H
8
)
=
44.1

g
m
o
l
−
1
, you can use this formula to determine (C3H8)
n
(
C
3
H
8
)
.

In this question it is quite hard to explain the use of significant figures. Those are used to imply a certain inaccuracy. Not enough information is given by the question, as of how accurate the measurement is. It is a mere exercise of converting one property into another. Here you should not worry about it.

5 0

3 years ago

Which two human activities are believed to be the greatest contributors to global warming?

vivado [14]

B. Deforestation and burning of fossil fuels.

7 0

3 years ago

Read 2 more answers

1. How many moles of nitrogen monoxide can be made using 5.0 moles of oxygen in

FrozenT [24]

Answer:

1) 10.0 moles of NO

2) 25 moles of NaCl

3) 1200 moles of CO2

4) 1.03 moles of MgO

5) 0.72 moles H2

6) 1041.15 grams BaCl2

7) 9.55 grams MgO

8) 45.5 grams Au

9 )14.93 grams AlCl3

Explanation:

1. How many moles of nitrogen monoxide can be made using 5.0 moles of oxygen in the following composition reaction?

N2 + O2 → 2NO

For 1 mol N2 we need 1 mol O2 to produce 2 moles of NO

For 5.0 moles of N2 we need 5.0 moles of O2 to produce 10.0 moles of NO

2. The neutralization of an acid with a base is a double replacement reaction in which a salt and water are formed. If you start with 25 moles of HCl and neutralize it with NaOH how many moles of NaCl will be formed?

HCl + NaOH → NaCl + H2O

For 1 mol HCl we need 1 mol NaOH to produce 1 mol of NaCl and 1 mol H2O

For 25 moles of HCl we need 25 moles of NaOH to produce 25 moles of NaCl and 25 moles of H2O

3. A car burns gasoline (octane – C8H18) with oxygen. If you drive to Salt Lake and burn 150 moles of octane how many moles of carbon dioxide are you producing?

2C8H18 + 25O2 → 16CO2 + 18H2O

For 2 moles of octane we need 25 moles of O2 to produce 16 moles of CO2 and 18 moles of H2O

For 150 moles of octane we need 25*75 = 1875 moles of O2

To produce 16*75 = 1200 moles of CO2 and 18*75= 1350 moles

4. If 25 gram of magnesium combines with oxygen in a composition reaction, how many moles of magnesium oxide will be formed?

2Mg + O2 → 2MgO

Moles of Mg = 25.0 g/24.3 g/mol = 1.03 moles

For 2 moles Mg we need 1 mol O2 to produce 2 moles MgO

For 1.03 moles Mg we'll have 1.03 moles of MgO

5. . Lithium reacts with water in a single replacement reaction. How many moles of hydrogen gas a produced by 10 grams of lithium?

2Li + 2H2O → 2LiOH + H2

Moles Li = 10.0 grams/ 6.94 g/mol = 1.44 moles

For 2 moles Li we need 2 mole H2O to produce 2 moles LiOH and 1 mol H2

For 1.44 moles Li we need 1.44 moles H2O to produce 1.44 moles H2O and 0.72 moles H2

6. Barium chloride reacts with sodium sulfate in a double replacement reaction. How many grams of barium chloride are required to react with 5 moles of sodium sulfate?

BaCl2 + Na2SO4 → BaSO4 + 2NaCl

For 1 mol of BaCl2 we need 1 mol of Na2SO4 to produce 1 mol of BaSO4 and 2 moles NaCl

For 5 moles Na2SO4 we need 5 moles BaCl2

mass BaCl2 = 5 moles * 208.23 g/mol = 1041.15 grams BaCl2

7. Magnesium carbonate when heated decomposes to form magnesium oxide and carbon dioxide. How many grams of magnesium oxide will be formed if 20 grams of magnesium carbonate are heated?

MgCO3 → MgO + CO2

Moles MgCO3 = 20.0 grams / 84.31 g/mol

Moles MgCO3 = 0.237 moles

For 1 mol MgCO3 we'll have 1 mol MgO and 1 mol CO2

For 0.237 moles MgCO3 we'll have 0.237 moles MgO and 0.237 moles CO2

Mass MgO = 0.237 moles * 40.30 g/mol = 9.55 grams MgO

8. If 70 grams of gold III chloride decomposes into its elements, how many grams of gold will be produced?

2AuCl3 → 2Au + 3Cl2

Moles AuCl3 = 70 grams / 303.33 g/mol = 0.231 moles

For 2 moles AuCl3 we'll have 2 moles gold and 3 moles Cl2

For 0.231 moles AuCl3 we'll have 0.231 moles gold

Mass of gold = 0.231 moles * 196.97 g/mol = 45.5 grams Au

9. Chlorine is more reactive element than bromine, thus chlorine will replace bromine in compound through a single replacement reaction. If 30 grams of aluminum bromide react with chlorine in this fashion how many grams of aluminum chloride will be formed?

2AlBr3 + 3Cl2 → 2AlCl3 + 3Br2

Moles AlBr3 = 30 g /266.69 g/mol = 0.112 moles

For 2 moles AlBr2 we need 3 moles Cl2 to produce 2 moles AlCl3 and 3 moles Br2

For 0.112 moles AlBr3 we need 3/2 * 0.112 = 0.168 moles of Cl2

To produce 0.112 moles of AlCl3 and 0.168 moles of Br2

Mass AlCl3 = 0.112 moles * 133.34 g/mol = 14.93 grams AlCl3

8 0

3 years ago

A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed

AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) Balanced equation including N_2 from air

The balanced equation ignoring N_2 from air is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2

Including N_2 from air, the balanced equation is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2

(b) Balanced equation for 120 % stoichiometric combustion

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) Minimum mass of air

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air

(d) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

Mass of air = 17 300 kg × 1.20 = 20 760 kg air

Air:fuel = 20 760 kg/1700 kg = 12.2 :1

6 0

3 years ago