CaCO3(s) ⟶ CaO(s)+CO2(s)
<span>
moles CaCO3: 1.31 g/100 g/mole CaCO3= 0.0131 </span>
<span>
From stoichiometry, 1 mole of CO2 is formed per 1 mole CaCO3,
therefore 0.0131 moles CO2 should also be formed.
0.0131 moles CO2 x 44 g/mole CO2 = 0.576 g CO2 </span>
Therefore:<span>
<span>% Yield: 0.53/.576 x100= 92 percent yield</span></span>
Answer is: 10 moles of water will be produced.
Balanced chemical reaction of formation of water:
2H₂ + O₂ → 2H₂O.
n(H₂) = 10 mol; amount of hydrogen gas.
From balanced chemical reaction: n(H₂) : n(H₂O) = 2 : 2 (1 : 1).
n(H₂O) = n(H₂).
n(H₂O) = 10 mol; amount of water.
Answer:
2,2,3,3-tetrapropyloxirane
Explanation:
In this case, we have to know first the alkene that will react with the peroxyacid. So:
<u>What do we know about the unknown alkene? </u>
We know the product of the ozonolysis reaction (see figure 1). This reaction is an <u>oxidative rupture reaction</u>. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If 
is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.
<u>What is the product with the peroxyacid?</u>
This compound in the presence of alkenes will produce <u>peroxides.</u> Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product <u>2,2,3,3-tetrapropyloxirane.</u> (see figure 2)
Equation of decomposition of ammonia:
N2+3H2->2NH3
Euilibrium constant:
Kc=(NH3)^2/((N2)((H2)^3))
As concentration of N2=0.000105, H2=0.0000542
so equation will become:
3.7=(NH3)^2/(0.000105)*(0.0000542)^3
NH3=√(3.7*0.000105*(0.0000542)^3)
NH3=7.8×10⁻⁹
So concentration of ammonia will be 7.8×10⁻⁹.
