Answer:
10.6 g CO₂
Explanation:
You have not been given a limiting reagent. Therefore, to find the maximum amount of CO₂, you need to convert the masses of both reactants to CO₂. The smaller amount of CO₂ produced will be the accurate amount. This is because that amount is all the corresponding reactant can produce before it runs out.
To find the mass of CO₂, you need to (1) convert grams C₂H₂/O₂ to moles (via molar mass), then (2) convert moles C₂H₂/O₂ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams (via molar mass). *I had to guess the chemical reaction because the reaction coefficients are necessary in calculating the mass of CO₂.*
C₂H₂ + O₂ ----> 2 CO₂ + H₂
9.31 g C₂H₂ 1 mole 2 moles CO₂ 44.0095 g
------------------ x ------------------- x ---------------------- x ------------------- =
26.0373 g 1 mole C₂H₂ 1 mole
= 31.5 g CO₂
3.8 g O₂ 1 mole 2 moles CO₂ 44.0095 g
------------- x -------------------- x ---------------------- x -------------------- =
31.9988 g 1 mole O₂ 1 mole
= 10.6 g CO₂
10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.
Answer:
There are
4.517
⋅
10
23
atoms of Zn in 0.750 mols of Zn.
Explanation:
Since we know that there are
6.022
⋅
10
23
atoms in every mole of a substance (Avogadro's Number), there are
6.022
E
23
⋅
0.750
atoms of Zn in 0.750 mols of Zn.
Answer:
So, 10 mole of water will weigh (18x10) = 180g.
Answer:
Metalloids
Explanation:
.... .....................
<span>8.21 L of C3H8(g)
Lets take c as the molar volume at that temperature.
c L <><> 5c L
C3H8 (g) + 5O2 (g) --> 3CO2 + 4H2O + Q
8.21 L <><> x L
x = (8.21 * 5c)/c = 8.21 * 5 = 41.05 L O2 consumed for a 100% yield.</span>
