The empirical and molecular formulas will be
and
respectively.
<h3>Empirical and molecular formula</h3>
The compound contains C, H, and O.
C = 61.15/12 = 5.0958
H = 5.3/1 = 5.3
O = 31.55/16 = 1.9719
Divide by the smallest
C = 2.6
H = 2.7
O = 1
Thus, the empirical formula is 
Empirical formula mass = (12x5) + (1x5) + 16x2 = 97
n = 152.15/97 = 2
The molecular formula is 
More on molecular and empirical formulas can be found here: brainly.com/question/14425592
#SPJ1
I believe it is 3 maybe
Hope I got it right
<u>Answer:</u> The correct answer is Option A.
<u>Explanation:</u>
Mole ratio is defined as the ratio between the stoichiometric coefficients of the molecules present in the chemical reaction.
For the given balanced chemical equation:

By Stoichiometry of the reaction:
3 moles of iron metal reacts with 4 moles of water to produce 1 mole of iron oxide and 4 moles of hydrogen gas.
The mole ratio of 
Hence, the correct answer is Option A.
Answer:
Change in enthalpy for the reaction is -536 kJ
Explanation:
- Overall chemical reaction can be represented a summation of two given elementary steps with slight modification.
- Take reaction (1a) and divide stoichiometric coefficients by 2
- Take reverse reaction (2a) and divide stoichiometric coefficient by 2
- Then add these two modified elementary steps to get overall chemical reaction
is an additive property. hence value of
will be changed in accordance with modification


--------------------------------------------------------------------------------------------------------------
