If there is 3 grams of carbon and 8 grams of oxygen how many grams of carbon dioxide is produced?

Answer 1

9, with two grams oxygen left over

Answer 2

We need a balanced reaction equation. In this case: C + O2 -> CO2. In terms of moles, 12 grams of carbon reacts with 32 grams of oxygen, so under the conditions stated, you have 0.25 mole of each and neither reactant is limiting. So, all of each reacts, and 11 grams of CO2 is produced.

Answer 3

Start with a balanced equation. C + O2 --> CO2. Divide the 3 g of carbon by it's molar mass to get how many moles of carbon there are. Divide the 8 g of O2 by it's molar mass to get the moles of oxygen available. If there is a limiting reactant, you will make only as much CO2 as that limiting reactant allows. If there is no limiting reactant, you will use all 3 g of C and 8 g of O2. By the way, Beardog's answer is incorrect-oxygen is diatomic (i.e. it's formula is O2, not O).

Answer 4

Let's assume the reaction works.

(1) Balance the equation:

C + O2 ---> CO2

(2) Determine the limiting reactant.

3 grams of C will produce 0.3 moles of CO2

8 grams of O2 will produce 0.25 moles of CO2.

Therefore, oxygen is the limiting reactant.

(3) Convert 0.25 moles to grams of CO2.

0.25 moles of CO2 * 44 g/mol CO2 = 11 grams CO2.

I didn't use significant digits, but you can do that.

Answer 5

11 grams of CO2

Answer 6

11 grams...0.75 mols of CO2

Answer 7

2C + 2O2 = 2CO2 volume of C = mass/molar mass = 3/12 =0.25mol comparing no. of mole of C to CO2, 2 mole of C provides 2 mole of CO2. subsequently mole ratio = a million in view that no. of mole = mass/molar mass, mass = no of mole x molar mass of CO2 = 0.25 x {12.0 + (sixteen.0 x 2) } = 0.25 x 40 4.0 = 11.0 g *molar mass of CO2 = Molar mass of a million C + Molar mass of two O

Answer 8

I keep trying to weigh it, but the oxygen keeps escaping!