I think the answer is C. 02
439.3 g CO2
Explanation:
First find the # of moles of CO2 that results from the combustion of 3.327 mol C3H6:
3.227 mol C3H6 × (6 mol CO2/2 mol C3H6)
= 9.981 mol CO2
Use the molar mass of CO2 to determine the # of grams of CO2:
9.981 mol CO2 x (44.01 g CO2/1 mol CO2)
= 439.3 g CO2
<u>Answer:</u> The percent yield of the compound is 30.86 %.
<u>Explanation:</u>
To calculate the percentage yield of a compound, we use the equation:

Experimental yield of compound = 25 g
Theoretical yield of compound = 81 g
Putting values in above equation, we get:

Hence, the percent yield of the compound is 30.86 %.
Answer:
Mass = 2.8 g
Explanation:
Given data:
Mass of HCl = 102 g
Mass of Zn = 128 g
Mass of H₂ = ?
Solution:
Chemical equation:
2HCl + Zn → H₂ + ZnCl₂
Number of moles of Zn:
Number of moles = mass/molar mass
Number of moles = 128 g/ 65.38 g/mol
Number of moles = 2 mol
Number of moles of HCl:
Number of moles = mass/molar mass
Number of moles = 102 g/ 36.5 g/mol
Number of moles = 2.8 mol
Now we will compare the moles of hydrogen gas with HCl and Zn.
HCl : H₂
2 : 1
2.8 : 1/2×2.8 = 1.4 mol
Zn : H₂
1 : 1
2 : 2
Number of moles of hydrogen formed by HCl are less thus it will limiting reactant.
Mass of Hydrogen:
Mass = number of moles × molar mass
Mass = 1.4 mol × 2 g/mol
Mass = 2.8 g