How many liters of O2 (g) are needed to react completely with 56.0 L of CH4 (g) at STP to produce CO2 (g) and H2O (g)? Given: CH4 (g) + 2O2 (g) → CO2 (g) + H2O (g)

Md Ariful Islam

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According to the equation 1L of CH4 reacts with 2L of O2.

Volume of O2 = V(CH4) *2 = 56*2 = 112 L

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CH4+2O2=CO2+2H2O HOW MANY MOLES OFOXYGEN ARE NEEDED TO REACT WITH 8.00 MOLES OF METHANE

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In the following reaction, how many liters of O<sub>2</sub> will produce 43.62 liters of CO<sub>2</sub> at STP? C<sub>3</sub>H<sub>8</sub> + 5 O<sub>2</sub> 3 CO<sub>2</sub> + 4 H<sub>2</sub>O

Litres of O2=72.7L O2

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Methane (CH4), the principal constituent of natural gas, reacts with oxygen according to this equation: CH4 + 2 O2 → CO2 + 2 H2O How many grams of O2 are needed to produce 37.5 g of CO2?

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131.3+-41.2-206.1= 116kJ

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How many carbon atoms are there on the reactant side of this equation CH4+2O2+--&gt; CO2 + 2H2O

Solution:The equation:CH4 + 2O2 = CO2 + 2H2OAccording to the equation: Number of reactants: 4 H, 1 C, 4 O.Number of products: 4 H, 1 C, 4 O.So, one carbon...

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CH4 + 2O2 ----&gt; CO2 + 2H2O Heat released is 890.4kJ How much heat is released by burning 15g of CH4?

Solution:The balanced thermochemical equation:CH4 + 2O2 → CO2 + 2H2O, △H = -890.4 kJ The equivalences for this thermochemical equation are:1 mol CH4 ⇔ 2 mol O2 ⇔ 1 mol CO2 ⇔ 2 mol...

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Q = n (-ΔH∘rxn) = m M-1 (-ΔH∘rxn) = 244 g (-802.3 kJ mol-1) /16.04 g mol-1 = 12205 kJ = 12.2 MJ Answer: 12.2 MJ

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The combustion of methane gas is described by the equation CH4+2O2→CO2+2H2O. What volume of oxygen is needed at STP to combust 4.7 L methane? How much carbon dioxide will be formed?

CH4+2O2→CO2+2H2O V(O_2 ):V(CH_4 )=2:1 V(O_2)=V(CH_4 )×2=4.7×2=9.4 L V(CO_2 ):V(CH_4 )=1:1 V(CO_2)=V(CH_4 )=4.7 L Answer: V(O2) = 9.4 L, V(CO2) = 4.7 L.

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The Principal constituent of natural gas is methane, which burns in air by this reaction: CH4 + 2O2 (arrow to the right) 2H2O + CO2 How many grams of O2 are needed to produce 3.5 gms of CO2?

Calculate number of moles of CO2 = mas s/ MW = 3.5 g / 44 g/mole = 0.08 moles Therefore, number of moles of O2= 2 x 0.08 moles =...

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V(O3)=5/2 = 2.5 L.

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