How many grams of water are produced from the combustion of 45.2 g of methane? *
CH4 + 202 → CO2 + 2H2O
O 90.4 g H20
O 2g H20
O 101.51 g H20
18.015 g H20

101.56 of H₂O

Explanation:

The balanced equation for the reaction is given below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

Next, we shall determine the mass of CH₄ that reacted and the mass of H₂O produced from the balanced equation. This is illustrated below:

Molar mass of CH₄ = 12 + (4×1.01)

= 12 + 4.04

= 16.04 g/mol

Mass of CH₄ from the balanced equation = 1 × 16.04 = 16.04 g

Molar mass of H₂O = (2×1.01) + 16

= 2.02 + 16

= 18.02 g/mol

Mass of H₂O from the balanced equation = 2 × 18.02 = 36.04g

SUMMARY:

From the balanced equation above,

16.04 g of CH₄ reacted to produce 36.04 g of H₂O.

Finally, we shall determine the mass of water, H₂O produced by the reaction of 45.2 g of methane, CH₄. This can be obtained as illustrated below:

From the balanced equation above,

16.04 g of CH₄ reacted to produce 36.04 g of H₂O.

Therefore 45.2 g of CH₄ will react to produce = (45.2 × 36.04)/16.04 = 101.56 g of H₂O.

Thus, 101.56 of H₂O were obtained.

1. they vary with a higher number because boiling is way hotter than melting

i would say b

explanation:

but i think i would need more information than that