Respuesta :
Answer:
[tex]m_{CaCO_3}=0.179gCaCO_3[/tex]
Explanation:
Hello,
In this case, since the undergoing chemical reaction is:
[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]
The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:
[tex]PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2[/tex]
Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:
[tex]m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3[/tex]
Regards.
The mass of CaCO₃ required to produce 40.0 mL of CO₂ at STP is 0.179 g
From the question,
We are to determine the mass of CaCO₃ required to produce 40.0 mL of CO₂ at STP.
First, we will determine the number of mole of CO₂ required to be produced
From the formula
PV = nRT
Where
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant
and T is the temperature
Then, we can write that
[tex]n = \frac{PV}{RT}[/tex]
From the question,
V = 40.0 mL = 0.04 L
At STP
P = 1 atm
T = 273.15 K
and
R = 0.08206 L atm mol⁻¹ K⁻¹
Putting the parameters into the formula, we get
[tex]n = \frac{1 \times 0.04}{0.08206 \times 273.15}[/tex]
∴ n = 0.0017845 mole
Now, we will write the balanced chemical equation for the decomposition of CaCO₃
CaCO₃ → CaO + CO₂
This means,
1 mole of CaCO₃ will decompose to produce 1 mole of CO₂
Since 0.0017845 mole of CO₂ is to be produced,
Then,
0.0017845 mole of CaCO₃ would be required
Now, for the mass of CaCO₃ required,
Using the formula
Mass = Number of moles × Molar mass
Molar mass of CaCO₃ = 100.0869 g/mol
∴ Mass of CaCO₃ required = 0.0017845 × 100.0869
Mass of CaCO₃ required = 0.178605 g
Mass of CaCO₃ required ≅ 0.179 g
Hence, the mass of CaCO₃ required to produce 40.0 mL of CO₂ at STP is 0.179 g
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