Problem page try again your answer is wrong. in addition to checking your math, check that you used the right data and did not round any intermediate calculations. gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . suppose 51. g of butane is mixed with 150. g of oxygen. calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. round your answer to significant digits.

The answer to your question is: 154.8 g of CO2

Explanation:

Butane = 51 g

Oxygen = 150 g

Carbon dioxide = ?

Reaction

                 2C₄H₁₀        +      13 O₂          ⇒        8CO₂       +         10H₂O    

MW           2(58)                   13(32)                     8( 44)                  10(18)

                   116g                    416g                       352g                  180g

LImiting reactant

                            116 g of butane 416 g oxygen

                             51 g                     x

                             x = (51 x 416) / 116

                             x = 182.9 g of oxygen

The limiting reactant is butane and excess reactant is oxygen because we need 182.9 g  and we have 416g.

Then we calculate the mass of Carbon dioxide with the mass of butane.

                           116 g of butane 352 g of CO2

                           51g                       x

                           x = (51 x 352) / 116

                           x = 154.8 g of CO2

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

what does the table look like