Answer to Question #155129 in General Chemistry for Jhocel Mae

Question

1. The combustion of acetylene gas is represented by the equation: 2C2H2(g)+ 5 O2(g) 4CO2(g) + 2H2O (g) . How many grams of carbon dioxide (CO2) and grams of water(H2O)are produced when 66.0 g of C2H2 burns in oxygen? 2. If there are 35.o grams of C6H10 and 45.0 grams of O2, how many grams of the excess reagent will remain after the reaction ceases? 6 C6H10 +17 O2 12 CO2 + 10 H2O 3. For the balanced equation below, if the reaction of 77.0 grams of CaCN2 produces 27.1 g of NH3, what is the percent yield? CaCN2 + 3 H2O CaCO3 + 2 NH3. I need an explanation to your answers everyone. Thank you so much.

Answer 1

1. M(C₂H₂) = 26 g/mol

m(C₂H₂) $= \frac{66.0}{26}=2.538 \;mol$

n(CO₂) = 2n(C₂H₂) $= 2 \times 2.538 = 5.076 \;mol$

M(CO₂) = 44 g/mol

m(CO₂) $= 44 \times 5.076 = 223.34 \;g$

n(H₂O) = n(C₂H₂) = 2.538 mol

M(H₂O) = 18 g/mol

m(H₂O) $= 18 \times 2.538 = 45.68 \;g$

2. M(C₆H₁₀) = 82 g/mol

n(C₆H₁₀) $= \frac{35.0}{82}=0.426 \;mol$

M(O₂) = 32 g/mol

n(O₂) $= \frac{45.0}{32}=1.406 \;mol$

Theoretical proportion:

C₆H₁₀ : O₂ = 6 : 17 = 1 : 2.83

Real proportion:

C₆H₁₀ : O₂ = 0.426 : 1.406 = 1 : 3.3

So, C₆H₁₀ is the limiting reactant.

According to the reaction:

n(O₂) $= \frac{17}{6}n(C_6H_{10}) = \frac{17}{6} \times 0.426 = 1.207 \;mol$

Used mass of O₂:

m’(O₂) $= 1.207 \times 32 = 38.624 \;g$

Δm(O₂) = 45.0 – 38.624 = 6.376 g

So, 6.376 g of O₂ (excess reagent) will remain after the reaction ceases.

3. M(CaCN₂) = 80 g/mol

n(CaCN₂) $= \frac{77.0}{80}= 0.9625 \;mol$

According to the reaction:

n(NH₃) = 2n(CaCN₂) $= 2 \times 0.9625 = 1.925 \;mol$

M(NH₃) = 17 g/mol

m’(NH₃) = $1.925 \times 17$ = 32.725 g (theoretical mass)

Proportion:

32.725 – 100 %

27.1 – x

$x =\frac{27.1 \times 100}{32.725} = 82.81\; \%$

So, he percent yield is 82.81 %.