(4.99 g) / (3.65 g/mL) = 1.37 mL
The answer to your question is 1.37ml.
Answer:
First confirm the reaction is balanced:
C3H8 + 5O2 --> 3CO2 + 4H20 (3 cabon - check; 8 hydrogen - check; 10 oxygen - check).
a) In the equation there is a 5:1 ratio between propane and oxygen. We also know that number of mole is proportional to pressure and volume. Since pressure is constant (STP) then the volume of O2 is 7.2 * 5 = 36 litres.
b) For a near ideal gas that PV = nRT (combined gas law). So for 7.2 litres propane we find n(propane) = 101.3 * 7.2/8.314*298 ~ 0.29 mole (using metric units throughout for simplicity).
There is a 1:3 ratio between propane and CO2. Therefore 3 * 0.29 = 0.87 mole of CO2 is produced.
MW(CO2) ~ 44 g/mol. Therefore m(CO2) = 44 * 0.87 ~ 38.3 g
c) We know we need more oxygen than propane (due to the 1:5 ratio) so oxygen is the limiting reagent. Again Volume is proportional to number of mole and we see there is a 5:4 ratio between oxygen and water. Therefore the volume of water vapour produced will be (4/5) * 15 = 12 litres.
The other questions use the same technique and will give you some much needed practice.
Explanation:
Answer:
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
Explanation:
Step 1: Data given
Mass of nitrogen gas (N2) = 13.4 grams
Molar mass of N2 = 28 g/mol
Molar mass of NH3 = 17.03 g/mol
Step 2: The balanced equation
N2 + 3H2 → 2NH3
Step 3: Calculate moles of N2
Moles N2 = Mass N2 / molar mass N2
Moles N2 = 13.4 grams / 28.00 g/mol
Moles N2 = 0.479 moles
Step 4: Calculate moles of NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles
Step 5: Calculate mass of NH3
Mass of NH3 = moles NH3 * molar mass NH3
Mass NH3 = 0.958 moles * 17.03 g/mol
Mass NH3 = 16.3 grams
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
