The grams that would be produced from 7.70 g of butanoic acid and excess ethanol is 7.923grams
calculation
Step 1: write the chemical equation for the reaction
CH3CH2CH2COOH + CH3CH2OH → CH3CH2CH2COOCH2CH3 +H2O
step 2: find the moles of butanoic acid
moles= mass/ molar mass
= 7.70 g/ 88 g/mol=0.0875 moles
Step 3: use the mole ratio to determine the moles of ethyl butyrate
moles ratio of CH3CH2CH2COOH :CH3CH2CH2COOCH2CH3 is 1:1 therefore the moles of CH3CH2CH2COOCH2CH3 = 0.0875 x78/100=0.0683moles
step 4: find mass = moles x molar mass
= 0.0683 moles x116 g/mol=7.923grams
Answer:
D. 5 moles
Explanation:
C3H8 + 5O2 → 4H2O + 3CO2
5 mol 3 mol
So, to make 3 mol CO2 5 mol O2 are needed.
The question is cut off in the picture
Combustion Have a great day!
Answer: D.
Explanation: D. Acetone, Acetone may be rinsed over materials to wash off the explosive residue.
Hope this helps :)
