Sample of 10.7g of O2 reacts completely with CO to form CO2

3a) How many grams of CO2 will be formed. So 10.7g (1mol/32g) (2molCO2/2molCO2)(22.01gCO2/1molCO2) = 14.7g. Please is this right
3b) is the mass of CO2 ypu determined in the equation (3a) the minimum or the maximum amout of CO2 produced in this question.
3c) How many grams of CO is required.
10.7g(1mol/32g)(2molCO/2molCO2)(28.01gCO/1molCO) = 8.8g
Please is due is this correct

i need help Drbob222

3a isn't correct. I didn't chck the others. It should be

10.7 x(1 mol O2/32g O2) x (2 mols CO2/1 mol O2) x (44 g CO2/mol CO2) = about 29 or so g CO2.
Check your factors. The units don't cancel as they should.

thank you can you check 3c, found my mistake on 3a

3c please is due now @DrBob222

Sorry to be so long but I had to leave home. The set up for 3c looks ok but I don't get that answer. Check your arithmetic. I get 9+ for an answer.

Yes, your calculations for questions 3a and 3c are correct.

3a) To find the mass of CO2 formed, you started with 10.7 grams of O2 and used the molar ratios from the balanced chemical equation to convert grams of O2 to grams of CO2. Your calculation correctly involves the molar mass of O2 (32g/mol) and CO2 (44.01g/mol), which is used to convert moles to grams.

10.7g (1mol/32g) (2molCO2/2molCO2)(44.01gCO2/1molCO2) = 14.7g

So, 14.7 grams of CO2 will be formed.

3b) The mass of CO2 you determined in equation 3a represents the maximum amount of CO2 that can be produced in this reaction. Since you started with a certain amount of O2, which is limiting in this reaction, it determines the maximum amount of CO2 that can be formed. It is important to note that the actual amount of CO2 formed may be less if there is not enough of the other reactant (CO) to completely react with all of the O2.

3c) To find the mass of CO required, you used the molar ratios from the balanced chemical equation to convert grams of O2 to grams of CO. Your calculation correctly involves the molar mass of O2 (32g/mol) and CO (28.01g/mol), which is used to convert moles to grams.

10.7g (1mol/32g) (2molCO/2molCO2)(28.01gCO/1molCO) = 8.8g

So, 8.8 grams of CO is required to completely react with 10.7 grams of O2.