Question

At stp how many liters of oxygen are required to react completely with 4.5 liters of hydrogen to form water? 2H2(g)+O2(g)—>2H2O(g)

1.8 l
3.0 l
9.0 l
2.0 l

how many moles of aluminum are needed to react completely with 4.9 mol of FeO? 2Al(s)+3FeO(s)—>3Fe(s)*Al2O3(s)
4.9 mol of Al
0.8 mol of Al
3.3 mol of Al
7.4 mol of Al

You are required to use the molar mass in all of the following stoichiometric calculations EXCEPT
volume-volume problems
mass-mass problems
mass-particle problems
mass-volume problems

Iron(lll) oxide is formed when iron combines with oxygen in the air. How many grams of Fe203 are formed when 23.7 g of Fe reacts completely with oxygen? 4Fe(s)+3O2(g)—> 2Fe2O3(s)
33.9
23.9
135
71.1

What is the mole ratio of D to B in the generic chemical reaction? 2A+B-> C+3D
1:1
2:1
3:2
3:1

How many grams of CO are needed to react with an excess of Fe203 to produce 108.5 g Fe? Fe2O3(s)+3CO(g)->3CO2(g)+2Fe(s)
157.8
81.6
36.3
325.5

Solid sodium reacts violently with water, producing heat, hydrogen gas, and sodium hydroxide. How many molecules of hydrogen gas are formed when 57.1 g of sodium are added to water? 2Na+2H2O->2NaOH+H2
2.4999
1.51*10^24
6.37*10^23
7.48*10^23

The masses of the reactants and the theoretical yield of the products for a chemical reaction are shown below. When this reaction was performed in the laboratory, the actual yield for Sbl3 was 2.4 g. What was the percent yield for Sbl3? 3Sb(1.2g) + 3I2(2.4g) ->2SbI3(3.2g) + Sb(0.4g)
25%
67%
75%
100%

Which conversion factor do you use first to calculate the number of grams of CO2 produced by the reaction of 50.6 g O2 with excess CH4? The equation for the complete combustion of methane is
CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)
44.0 g CO2/2 mol CO2
1 mol O2/32.0 g O2
2 mol O2/1 mol CO2
1 mol CH4/16.0 g CH4

Explain the significance of coefficients in a balanced chemical reaction.

Answer 1

Do you have the answers i need help

Answer 2

1) The answer is: 3.0 liters of oxygen are required.

Balanced chemical reaction: 2H₂(g) + O₂(g) → 2H₂O(g).

V(H₂) = 4.5 L; volume of hydrogen.

Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).

At STP one mole of gas occupies 22.4 liters of volume.

In this question all substances are gases, amount of substance depends on volume of the gas.

V(H₂) : V(O₂) = n(H₂) : n(O₂).

4.5 L : V(O₂) = 2 mol : 1 mol.

V(O₂) = 2.25 L; volume of oxygen.

2) The answer is: 3.3 moles of aluminium are needed.

Balanced chemical reaction: 2Al(s) + 3FeO(s) → 3Fe(s) + Al₂O₃(s).

n(FeO) = 4.9 mol.

From balanced chemical reaction: n(FeO) : n(Al) = 3 : 2.

n(Al) = 2 · n(FeO) ÷ 3.

n(Al) = 2 · 4.9 mol ÷ 3.

n(Al) = 3.27; amount of aluminium.

3) The answer is: volume-volume problems.

For volume-volume problems, amount of substance is needed, not molar mass.

The base SI unit for molar mass is kg/mol, but chemist more use g/mol (gram per mole).

For example, the molar mass of fluorine is 38.00 g/mol.

M(F₂) = 2 · Ar(F) · g/mol.

M(F₂) = 2 · 19.00 · g/mol.

M(F₂) = 38.00 g/mol.

Molar mass M represent the mass of a substance (in this example molecule of florine) divided by the amount of substance.

4) The answer is: 33.9 grams of iron(III) oxide.

Balanced chemical reaction: 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s).

m(Fe) = 23.7 g; mass of iron.

n(Fe) = m(Fe) ÷ M(Fe).

n(Fe) = 23.7 g ÷ 55.85 g/mol.

n(Fe) = 0.424 mol; amount of iron.

From chemical reaction: n(Fe) : n(Fe₂O₃) = 4 : 2 (2 : 1).

n(Fe₂O₃) = 0.212 mol; amount of iron(III) oxide.

m(Fe₂O₃) = 0.212 mol · 159.69 g/mol.

m(Fe₂O₃) = 33.88 g; mass of iron(III) oxide.

5) The answer is: the mole ratio of D to B is 3 : 1.

Balanced chemical reaction: 2A + B → C + 3D.

Coefficients with the lowest ratio indicate the relative amounts of substances in a reaction.

Coefficient in fron of D is 3 and coefficient in fron B is 1.

A and Bare reactants and C and D are are products of this chemical reaction.

6) The answer is: 81.6 grams of CO are needed.

Balanced chemical reaction: Fe₂O₃(s) + 3CO(g) ⟶ 2Fe(s) + 3CO₂(g).

m(Fe) = 108.5 g; mass of iron.

n(Fe) = m(Fe) ÷ M(Fe).

n(Fe) = 108.5 g ÷ 55.85 g/mol.

n(Fe) = 1.94 mol; amount of iron.

From chemical reaction: n(Fe) : n(CO) = 2 : 3.

n(CO) = 3 · 1.94 mol ÷ 2.

n(CO) = 2.91 mol; amount of carbon(II) oxide.

m(CO) = 2.91 mol · 28 g/mol.

m(CO) = 81.6 g; mass of carbon(II) oxide.

7) The answer is: 7.48*10^23 of hydrogen gas molecules.

Balanced chemical reaction: 2Na + 2H₂O → 2NaOH + H₂.

m(Na) = 57.1 g; mass of sodium.

n(Na) = m(Na) ÷ M(Na).

n(Na) = 57.1 g ÷ 23 g/mol.

n(Na) = 2.48 mol.; amount of sodium.

From chemical reaction: n(Na) : n(H₂) = 2 : 1.

n(H₂) = 1.24 mol; amount of hydrogen.

N(H₂) = n(H₂) · Na.

N(H₂) = 1.24 mol · 6.022·10²³ 1/mol.

N(H₂) = 7.48·10²³; number of hydrogen molecules.

8) The answer is: the percent yield for Sbl3 is 75%.

Balanced chemical reaction: 3Sb + 3I₂ → 2SbI₃ + Sb.

m(SbI₃) = 2.4; actual mass of antimony(III) iodide.

m(SbI₃) = 3.2 g; theoretical mass of antimony(III) iodide.

Percent yield = actual yield / theoretical yield.

Percent yield = 2.4 g ÷ 3.2 g · 100%.

Percent yield = 75%.

9) The answer is: first conversion factor is 1 mol O2/32.0 g O2.

Balanced chemical reaction of methane combustion:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g).

m(O₂) = 50.6 g; mass of oxygen.

n(O₂) = m(O₂) ÷ M(O₂).

M(O₂) = 32.0 g/mol; molar mass of oxygen.

First, amount of oxygen must be calculated.

n(O₂) = 50.6 g ÷ 32 g/mol.

n(O₂) = 1.58 mol; amount of oxygen.

10) The answer is: Coefficients are important to prove the law of conservation of mass.

For example, balanced chemical reaction:

CaCl₂ + 2AgNO₃ → Ca(NO₃)₂ + 2AgCl.

Number of atoms must be the same on left and right side of balanced chemical reaction according to the law of conservation of mass or principle of mass conservation.

There are two chlorine atoms (Cl), one calcium atom (Ca), six oxygen atoms (O), one nitrogen atoms (N) and two silver atoms (Ag) in this balanced chemical reaction.