Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 1.3 g of methane is mixed with 2.30 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

1.29 g of H₂O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CH₄ + 2O₂ –> CO₂ + 2H₂O

Next, we shall determine the masses of CH₄ and O₂ that reacted and the mass of H₂O produced from the balanced equation. This can be obtained as follow:

Molar mass of CH₄ = 12 + (4×1) = 12 + 4 = 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of O₂ = 16 × 2 = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

Molar mass of H₂O = (2×1) + 16 = 2 + 16 = 18 g/mol

Mass of H₂O from the balanced equation = 2 × 18 = 36 g

SUMMARY:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂ to produce 36 g of H₂O.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂.

Therefore, 1.3 g of CH₄ will react with = (1.3 × 64)/16 = 5.2 g of O₂.

From the calculations made above, we can see that it will take a higher mass (i.e 5.2 g) of O₂ than what was given (i.e 2.30 g) to react completely with 1.3 g of CH₄.

Therefore, O₂ is the limiting reactant and CH₄ is the excess reactant.

Finally, we shall determine the maximum mass of water (H₂O) that can be obtained from the reaction.

To obtain the maximum mass of H₂O, the limiting reactant (i.e O₂) will be use as follow:

From the balanced equation above,

64 g of O₂ reacted to produce 36 g of H₂O.

Therefore, 2.3 g of O₂ will react to produce = (2.3 × 36)/64 = 1.29 g of H₂O.

Thus, 1.29 g of H₂O is obtained from the reaction.

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the answer is c. nacl