The decomposition of limestone CaCO3 > CaO + CO2
Answer:
16.9g of H₂O can be formed
Explanation:
Based on the chemical reaction, 2 moles of H₂ react per mole of O₂. To anser this question we must find limiting reactant converting the mass and volume of each reactant to moles:
<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>
8.76g * (1mol / 2.016g) = 4.345 moles
<em>Moles O₂:</em>
PV = nRT
PV/RT = n
P = 1atm at STP
V = 10.5L
R = 0.082atmL/molK
T = 273.15K at STP
n = 1atm*10.5L / 0.082atmL/molK*273.15K
n = 0.469 moles of oxygen
For a complete reaction of 4.345 moles moles of hydrogen are required:
4.345 moles H2 * (1mol O2 / 2mol H2) = 2.173 moles of O2 are required. As there are just 0.469 moles, Oxygen is limiting reactant
Now, 1 mole of O2 produce 2 moles of H2O. 0.469 moles will produce:
0.469 moles O₂ * (2 moles H₂O / 1mol O₂) = 0.938 moles H₂O.
The mass is -Molar mas H₂O = 18.01g/mol-:
0.938 moles * (18.01g/mol) =
<h3>16.9g of H₂O can be formed</h3>
Based on the diagram shown, a numerical setup for calculating the gram-formula mass for reactant 1 would be :
6(1) + 2(12) + 16
Hope this helps
Answer: I believe the correct answer would be A.
