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1. If you have 185 mol of CH4, how many moles of H2O would that produce? CH4 + 2 O2 🡪 CO2 + 2 H2O
Moles of H2O = 185 × 2 = 370 moles
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If 85.0 L of natural gas, which is essentially methane (CH4), undergoes complete combustion at STP, how many Liters of CO2 is produced CH4 + O2 → CO2+ H2O
From the equation 1 mole of CH4 produces 1 mole of CO2 Molar Mass of CH4 = 16.04 g/mol= 85 L = 85000 g / 16.04= 5299.252 moles Molar Mass...
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A mixture of 206.0 g 206.0 g of Cl2 and 45.0 g 45.0 g of P P reacts completely to form PCl3 and PCl 5 . PCl5. Find the mass of PCl5 produced.
Mass of PCl5 produced=129.11g
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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?
Grams of O2=43.2g
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Estimate the amount of heat released (in kj/mol) when 10g of CH4 gas reacts with an excess of O2 gas to form gaseous CO2 and liquid water. The molar mass of CH4 is 16.042
CH4(g) + 2O2(g) = CO2(g) + 2H2O (liq)∆H = 2∆H(H2O) + ∆H(CO2) - ∆H(CH4) ∆H = 2 * (-285,8) +(-393,5) + 74,6 = -890,5 kj/mol Q = n * ∆H;...
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CH4 + 2O2 ----> 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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Number of atoms before reaction CH4+2 O2 ——> CO2 + 2 H2O Number of atoms after reaction CH4+2 O2 ——> CO2 + 2 H2O
Solution:CH4 + 2O2 → CO2 + 2H2ONumber of atoms before reaction: One carbon atom (C); four hydrogen atoms (H); four oxygen atoms (O).Hence, 9 atoms before reaction.Number of atoms after...
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CH4 + O2 ⇌ H2CO + H2O
initial Kc
CH4 1.5895 0.6545
O2 2.018665
H2CO 0.00
H2O 0.00 what are the rest of the Kc values
Kc= 0.978857211
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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)
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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