Question

help with my study guide please

us He will react with gaseous oxygen (O2) to produce gaseous carbon dioxide 2) and gaseous water (H2O). Suppose 1.12 g of methane is mixed with 1.7 g on Oxygen Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits. CH_(g) + O2(g) → _CO2(g) +_ H2O(g)

Answer 1

Balanced chemical equation is:

CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(g)

Molar mass of CH4,

MM = 1*MM(C) + 4*MM(H)

= 1*12.01 + 4*1.008

= 16.042 g/mol

mass(CH4)= 1.12 g

use:

number of mol of CH4,

n = mass of CH4/molar mass of CH4

=(1.12 g)/(16.04 g/mol)

= 6.982*10^-2 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 1.7 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(1.7 g)/(32 g/mol)

= 5.312*10^-2 mol

1 mol of CH4 reacts with 2 mol of O2

for 6.982*10^-2 mol of CH4, 0.1396 mol of O2 is required

But we have 5.312*10^-2 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

According to balanced equation

mol of CO2 formed = (1/2)* moles of O2

= (1/2)*5.312*10^-2

= 2.656*10^-2 mol

use:

mass of CO2 = number of mol * molar mass

= 2.656*10^-2*44.01

= 1.169 g

Answer: 1.2 g