Answer:
The molecular formula of the compound is
. The molecular formula is obtained by the following expression shown below

Explanation:
Given molecular mass of the compound is 176 g/mol
Given empirical formula is
Atomic mass of carbon, hydrogen and oxygen are 12 u , 1 u and 16 u respectively.
Empirical formula mass of the compound = 


Molecular formula = 4
Molecular formula is 
Answer:
1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.
2) When 10.4 grams of carbon monoxide(g) react with excess water(l), 1.04 kJ of energy are absorbed.
Explanation:
1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).
2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ
The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:

2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).
CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g) ΔH = 2.80 kJ
The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is:

The volume of CO2 at STP =124.298 L
<h3>Further explanation</h3>
Given
Reaction
4 KMnO4, +4 C3H5(OH)5, -7K2CO3, + 7 Mn2O3, +5 CO2, + 16 H2O
701,52 g of KMnO4
Required
volume of CO2 at STP
Solution
mol KMnO4 (MW=158,034 g/mol) :
mol = mass : MW
mol = 701.52 : 158.034
mol = 4.439
mol CO2 from equation : 5/4 x mol KMnO4 = 5/4 x 4.439 = 5.549
At STP 1 mol = 22.4 L, so for 5.549 moles :
=5.549 x 22.4
=124.298 L
Answer:
27%
Explanation:
Hello,
The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

Finally, the percent yield turns out into:

%
Best regards.