First, we have to get the theoretical yield of CaO:
the balanced equation for the reaction is:
CaCO3(s)→CaO(s) +CO2(g)
covert mass to moles:
moles CaCO3 = mass of CaCO3 / molar mass of CaCO3
= 2x10^3 /100 = 20 moles
the molar ratio between CaCO3 : CaO = 1:1
∴moles of CaO = 1* 20 = 20 moles
∴mass of CaO = moles of CaO * molar mass of CaO
= 20 * 56 = 1120 g
∴the theoritical yield = 1120 g and we have the actual yield =1.05X10^3
∴Percent yield = actual yield / theoritical yield *100
= (1.05x10^3) / 1120 * 100
= 94 %
That looks like C8H13NO3 : Nitrocyclooctanone
Answer:
1. 26.30 g C.
2. 4.42 g H.
3. 35.04 g O.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the required as follows:
1. Here, the only source of carbon is in CO2, and thus, we calculate the grams of carbon from the produced grams of this substance:
2. Here, the only source of hydrogen is in H2O, and thus, we calculate the grams of hydrogen from the produced grams of this substance:
3. Here, we subtract the mass of H and C from the mass of the sample, to obtain the mass of oxygen:
Regards!
Density (kg/m³) = mass (kg) / Volume (m³)
d = m/V (1)
Ideal gas law,
PV = nRT (2)
<span>Where, P is the
pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the
universal gas constant ( 8.314 J mol</span>⁻¹ K⁻<span>¹) and T is temperature in Kelvin.
</span>
n = m/M (3)
Where, n is number of moles, m is mass and M is molar mass.
From (2) and (3),
PV = (m/M) RT
By rearranging,
P = (m/VM)RT (4)
From (1) and (4)
P = (dRT) / M
P = standard pressure = 1 atm = 101325 pa
d = ?
R = 8.314 J mol⁻¹ K⁻¹
T = Standard temperature = 273 K
M = 44 g/mol = 44 x 10⁻³ kg/mol
By substitution,
101325 Pa = (d x 8.314 J mol⁻¹ K⁻¹ x 273 K) / 44 x 10⁻³ kg/mol
d = (101325 Pa x 44 x 10⁻³ kg/mol) / (8.314 J mol⁻¹ K⁻¹ x 273 K)
d = 1.96 kg m⁻³ = 1.96 g/L
Hence, the density of the CO₂ at STP is 1.96 g/L
Assumption made is "CO₂ gas has an ideal gas behavior".