Answer:
%N = 25.94%
%O = 74.06%
Explanation:
Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅
We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.
m(N): 2 × 14.01 g = 28.02 g
Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅
We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.
m(O): 5 × 16.00 g = 80.00 g
Step 3: Calculate the mass of 1 mole of N₂O₅
We will sum the masses of N and O.
m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g
Step 4: Calculate the percent composition of N₂O₅
We will use the following expression.
%Element = m(Element)/m(Compound) × 100%
%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%
%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%
8.5L H2O (1 mol H2O/22.4L)=.37946 mol H2O
.3796 mol H2O(1 mol CH4/2 mol H2O)=.18973 mol CH4
.18973 mol CH4(22.4L/1mol CH4)=4.25L CH4 gas
Answer:3.7427017316233497
Explanation:
You must burn 1.17 g C to obtain 2.21 L CO₂ at STP.
The balanced chemical equation is
C + O₂ → CO₂.
<em>Step 1</em>. Convert <em>litres of CO₂ to moles of CO₂</em>.
STP is <em>0 °C and 1 bar</em>. At STP the volume of 1 mol of an ideal gas is 22.71 L.
Moles of CO₂ = 2.21 L CO₂ × (1 mol CO₂/22.71 L CO₂) = 0.097 31 mol CO₂
<em>Step 2</em>. Use the molar ratio of C:CO₂ to <em>convert moles of CO₂ to moles of C
</em>
Moles of C = 0.097 31mol CO₂ × (1 mol C/1 mol CO₂) = 0.097 31mol C
<em>Step 3</em>. Use the molar mass of C to <em>calculate the mass of C
</em>
Mass of C = 0.097 31mol C × (12.01 g C/1 mol C) = 1.17 g C
It looks as if you are using the <em>old (pre-1982) definition</em> of STP. That definition gives a value of 1.18 g C.