Brainliesttt !me : )

when methane, ch4, is combusted, it produces carbon dioxide, co2, according to the unbalanced equation: ch4 + o2 → co2 + h2o.

write the balanced equation for this reaction, and explain how it is possible for 10 grams of methane fuel to burn and emit 27 grams of carbon dioxide. discuss whether or not this reaction obeys the law of conservation of mass.

the balanced equation is CH4 + 2O2 -> CO2 + 2H20

moles = mass ÷ mr

moles of methane = 10 ÷ 16

= 10/16

= 5/8 moles

the moles of carbon dioxide are the same as methane because the ratio is 1:1

mass = moles × Mr

mass of carbon dioxide = 5/8 × 44

= 220 ÷ 8

= 27.5 grams

to find out whether it obeys the law of conservation of mass, you must find the mass of the reactants and the mass of the products. if they are the same, it obeys the law of conservation of mass. otherwise it doesnt.

the mass of oxygen is 32 × 5/8

= 160/8

= 20g

we already have the mass of methane (10g) so add it on

mass of reactants = 20 + 10 = 30g

find mass of water: 5/8 × 18

= 90/8

= 11.25g

mass of products = 11.25 + 27.5 = 38.75g

the mass of reactants vs products are not the same so it does not obey the law of conservation of mass

3 1/4 %

explanation: