Name Date Consider the following Reaction. A chemist allows 23.2g of CH4 and 78.
ID: 559779 • Letter: N
Question
Name Date Consider the following Reaction. A chemist allows 23.2g of CH4 and 78.3g O2 to react. When the reaction is finished, the chemist collects 52.7g CO2. Determine the limiting reagent, theoretical yield, and percent yield for the reaction. [Hint: first step balance the Chemical Reaction] Consider the following Reaction. A chemist allows 30.6g of Mg(OH)2 and 63.6g H3PO4 to react. When the reaction is finished, the chemist collects 34.7g Mgs(PO)2 Determine the limiting reagent, theoretical yield, and percent yield for the reaction. [Hint: first step balance the Chemical Reaction]Explanation / Answer
1)
a)
Molar mass of CH4 = 1*MM(C) + 4*MM(H)
= 1*12.01 + 4*1.008
= 16.042 g/mol
mass of CH4 = 23.2 g
we have below equation to be used:
number of mol of CH4,
n = mass of CH4/molar mass of CH4
=(23.2 g)/(16.042 g/mol)
= 1.446 mol
Molar mass of O2 = 32 g/mol
mass of O2 = 78.3 g
we have below equation to be used:
number of mol of O2,
n = mass of O2/molar mass of O2
=(78.3 g)/(32 g/mol)
= 2.447 mol
we have the Balanced chemical equation as:
CH4 + 2 O2 ---> CO2 + 2 H2O
1 mol of CH4 reacts with 2 mol of O2
for 1.4462 mol of CH4, 2.8924 mol of O2 is required
But we have 2.4469 mol of O2
so, O2 is limiting reagent
b)
we will use O2 in further calculation
Molar mass of CO2 = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol
From balanced chemical reaction, we see that
when 2 mol of O2 reacts, 1 mol of CO2 is formed
mol of CO2 formed = (1/2)* moles of O2
= (1/2)*2.4469
= 1.223 mol
we have below equation to be used:
mass of CO2 = number of mol * molar mass
= 1.223*44.01
= 53.84 g
Answer: theoretical yield = 53.8 g
% yield = actual mass*100/theoretical mass
= 52.7*100/53.84
= 97.9 %
Answer: 97.9 %
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