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The compound contains Carbon, Hydroxide and Oxide
1 mole of carbon iv oxide contains 44 g, out of which 12 g are carbon.
Therefore, 1.6004 g of CO2 will contain;
1.6004 ×12/44 = 0.4365 g of carbon
1 mole of water contains 18 g of which 2 g is hydrogen,
Therefore, 0.6551 g of H2O will hace ;
0.6551 × 2/18 = 0.0728 g of hydrogen.
The total mass of the compound is 0.8009 g,
Thus the mass of oxygen = 0.8009 -(0.4365 +0.0728)
= 0.2916 g
To get the empirical formula we first get the number of moles of each element;\
Carbon = 0.4365/12= 0.036375 moles
Hydrogen = 0.0728/1 = 0.0728 moles
Oxygen = 0.2916/16 = 0.018225 moles
Then, to get the smallest ratio we divide each with the smallest value;
Carbon : Hydrogen : Oxygen
= (0.036375/0.018225) : (0.0728/0.018225) : ( 0.018225/0.018225)
= 1.996 : 3.995 : 1
≈ 2 : 4 : 1
Therefore, the empirical formula is C2H4O
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Next, we compute the reacting moles of each reactants:
Then, as magnesium and hydrohloric acid are in a 1:2 molar ratio 0.153 moles of magnesium will completely react with 0.306 moles of hydrochloric acid yet we only have 0.144 moles, therefore, limiting reactant is hydrochloric acid. Thus, we compute the produced moles of hydrogen:
Finally, we use the ideal gas equation with T=298K and 1atm (STP conditions) to compute the liters of hydrogen gas:
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