Calculate the amount of carbon dioxide that can could produced when 1mole of carbon is burnt in air& 1mole is burnt in 16g of dioxygen &2 moles of carbon are burnt in16g of dioxygen

Answer 1

First, write balanced equations:

C + O2 → CO2

16gO2 x 1 mole O2/32 gO2 = (1/2) mole O2

1) So, 1 mole C in air (O2) will yield 1 mole CO2, if air (O2) is not limiting.

2) If you have only a 1/2 mole of O2, then, O2 is limiting and you will produce 1/2 mole CO2.

3) If you have 1/2 mole of O2 and 2 moles of C, then O2 is still limiting and you can only produce 1/2 mole CO2.

Answer 2

Calculate the amount of CO₂ that is produced when a)1 mol C is burned in air, b)1 mol C is burned in 16 g O₂, and c)2 mol C are burned in 16 g O₂.

This is a stoichiometry problem, so you must have a balanced equation. I generally write the equation, balance it, then write the quantities (the given one(s) and use ? for what you are being asked) below the substances, and write the the molar masses (from the periodic table) above the substances.

a) 12 g/mol 32 g/mol 44 g/mol (molar masses)

C + O₂ —→ CO₂ (balanced, no coefficients needed)

1 mol ? (given, asked)

To do stoichiometry, you will use the mol ratio of two substances in the balanced equation, such as these

examples:

1 mol CO₂ 1 mol O₂ 1 mol CO₂ 1 mol C

1 mol C 1 mol C 1 mol O₂ 1 mol O₂

In this problem, the conversion factor will be a ratio of a reactant to a product , such as: 1 mol CO₂

1 mol C

Use dimensional analysis with the conversion factor from the balanced equation:

1 mol C x 1 mol CO₂ = 1 mol CO₂ (which can be converted from mol → g , if needed)

1 mol C

b) In this next case, you will have an one more calculation step (a conversion), in addition to that shown

above:

12 g/mol 32 g/mol 44 g/mol (molar masses)

C + O₂ —→ CO₂ (balanced equation)

1 mol 16 g ? (given, asked)

You already determined that 1 mol of C (with excess O₂) will produce 1 mol CO₂, so now you must determine how much CO₂ will be produced from 16 g O₂. Whichever reactant, 1 mol C or 16 g O₂, makes the least amount of product is the limiting reactant, which determines how much product is made from the quantities given.

Convert the O₂ quantity from g → mol, then do the stoichiometry:

16 g O₂ x 1 mol O₂ = 0.50 mol O₂

32 g O₂

0.5 mol O₂ x 1 mol CO₂ = 0.5 mol CO₂ (which can be converted from mol → g , if needed)

1 mol O₂

Since this is less than the amount of product you can make from 1 mol C, the O₂ is the limiting reactant upon which any calculations must be based.

Therefore, given 1 mol C and 16 g O₂ you can make 0.5 mol CO₂.

c) For the last part of the problem, you must determine how much product you can make using 2 mol C

and compare that calculated quantity with the previous calculation based on 16 g O₂.

12 g/mol 32 g/mol 44 g/mol (molar masses)

C + O₂ —→ CO₂ (balanced equation)

2 mol 16 g ? (given, asked)

2 mol C x 1 mol CO₂ = 2 mol CO₂

1 mol C

From 2 mol C you can make 2 mol CO₂ and from 16 g O₂ you can make 0.5 mol CO₂ ; therefore, the O₂ is the limiting reactant, and the calculated amount of product must be based on the quantity of O₂ given, not the quantity of C given, which you did in the previous part of this problem.

From 2 mol C and 16 g O₂, therefore, you can produce 0.5 mol CO₂.