Use the equation for the combustion of butane: 2 C, H, + 13 0,8 CO, +10 H, O 4. If you had 99.08g of butane react, how many grams of carbon dioxide would be produced? 5. If you had 290.38 grams of butane, how many grams of water would be produced? 6. How many grams of oxygen gas are needed for the complete combustion of 307.47 grams of butane? 7. How many grams of oxygen would be required if you had 160.87 grams of butane?​

See explanation

Explanation:

Equation of the reaction;

C4H10 + 13/2 O2> 4CO2  + 5H2O

Number of moles of Butane = 99.08g/58 g/mol = 1.71 moles

1 mole of Butane yields 4 moles of CO2

1.71 moles of Butane yields 1.71 * 4 = 6.84 moles Of CO2

Mass of CO2 = 6.84 moles Of CO2 * 44 g/mol = 300.96 g

5. Number of moles of butane = 290.38 g/58 g/mol = 5 moles

1 mole of butane yields 5 moles of water

5 moles of butane yields 5 * 5 = 25 moles of water

Mass of water = 25 moles of water * 18 g/mol = 450 g of water

6. Number of moles of butane = 307.47g/58 g/mol = 5.3 moles

1 mole of butane reacts with 6.5 moles of oxygen

5.3 moles of butane reacts with 5.3 * 6.5 = 34.45 moles of oxygen

Mass of oxygen = 34.45 moles of oxygen * 32 g/mol = 1102.4 g

7.  Number of moles of butane = 160.87/58 g/mol = 2.8 moles

1 mole of butane reacts with 6.5 moles of oxygen

2.8 moles of butane reacts with 2.8 * 6.5 = 18.2 moles of oxygen

Mass of oxygen = 18.2  moles of oxygen * 32 g/mol = 582.4 g

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !