How many grams of oxygen are required to completely burn 85.6 grams of methane (CH4)?

CH4 + O2 → H2O + CO2

oops: O2 = 32, not 16, so that's 170.78g

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To determine the number of grams of oxygen required to completely burn 85.6 grams of methane (CH4), we need to use stoichiometry.

First, we need to balance the chemical equation:
CH4 + 2O2 → 2H2O + CO2

The balanced equation tells us that for every one mole of CH4, we need two moles of O2 to completely burn it.

Next, we need to calculate the number of moles of CH4:
Mass of CH4 = 85.6 grams
Molar mass of CH4 = 12.01 g/mol (carbon) + 4 * 1.01 g/mol (hydrogen) = 16.05 g/mol
Moles of CH4 = mass of CH4 / molar mass of CH4
= 85.6 g / 16.05 g/mol
≈ 5.33 mol

Since we need two moles of O2 for every one mole of CH4, we multiply the moles of CH4 by the stoichiometric coefficient:
Moles of O2 = 5.33 mol CH4 * 2 mol O2 / 1 mol CH4
= 10.66 mol O2

Finally, we calculate the mass of O2 required:
Mass of O2 = moles of O2 * molar mass of O2
= 10.66 mol * 32.00 g/mol (molar mass of O2)
≈ 341.12 grams

Therefore, approximately 341.12 grams of oxygen are required to completely burn 85.6 grams of methane.

85.6g CH4 = 5.34 moles

each mole CH4 reacts with 1 mole of O2
so, you need 5.34 moles O2 = 85.39g O2

How many grams of 02 are needed to form 17 grams of H2O

n=M/RFM

n=number of moles RFM= relative formula mass

n=17/18=0.944
0.944 moles
then
0.944=x/16
x=0.944*16
x=15.104
15.105 grams