Answer:
The mass of C2H2 in the mixture is 0.56gram using the ratio of carbon in the products contributed by the C2H2.
Explanation:
The balanced equation for the reaction is: C3H8 + 2C2H2 + 10O2 >> 7CO2 + 6H2O.
From the reaction, we know that the oxygen was in excess, this will make the Carbon sources the limiting agents in the reaction. The details of the reaction showed that the ratio of water to the carbon dioxide is 1.6:1. This also means that the expected mole of carbon dioxide will be 7/1.6, which is 3.75moles.
The individual balanced equation of reaction is:
C3H3 +5O2 >> 3CO2 + 4H2O
and 2C2H2 + 5O2 >>4CO2 + 2H2O. From this one can quickly tell that the propane is in sufficient supply as it produces 3 moles of CO2 out of the expected 3.75 moles obtained above. Leaving 0.75moles of CO2 to the ethyne.
The mass of ethyne in the mixture will therefore be: 0.75/3.75 X 2.8 = 0.56g.
The answer is 40 grams. If gold is 20 grams per mL and you have 2 mLs, you get 40 grams of gold. Just add 20 + 20 or multiply 2 x 20 to get 40.
Answer:
B: Na(s) + Cl2(g) + 3O2(g) = 2NaClO3(s)
Explanation:
We are looking for enthalpy of formation, so we want to see reactance in their natural standard form.
Thus, we want to see the reactance of Na, Cl2 and O2.
The only option that has the correct form of Na, Cl2 and O2 is B.
Na(s) + Cl2(g) + 3O2(g) = 2NaClO3(s)
