<span>4 C3H5(NO3)3 = 12 CO2 + 6 N2 + 10 H2O + O<span>2
That's the answer lovely~</span></span>
Answer:
5.66 %.
Explanation:
<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>
<em />
<em>mass % = (mass of solute/mass of solution) x 100.</em>
<em></em>
mass of potassium nitrite = 30.0 g,
mass of the solution = mass of water + mass of potassium nitrite = 500.0 g + 30.0 g = 530.0 g.
<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (30.0 g/530.0 g) x 100 = <em>5.66 %.</em>
Answer:
The volume of nitrogen oxide formed is 35.6L
Explanation:
The reaction of nitric acid with copper is:
Cu(s) + 4HNO₃ → Cu(NO₃)₂ + 2NO₂(g) + 2H₂O(l)
Moles of copper are:

Moles of nitric acid are:

As 1 mol of Cu reacts with 4 moles of HNO₃:
0.697 mol Cu × (4mol HNO₃ / 1mol Cu) = 2.79 moles of HNO₃ will react. That means Cu is limiting reactant.
Moles of NO₂ produced are:
0.697 mol Cu × (2mol NO₂ / 1mol Cu) = <em>1.394 moles of NO₂</em>
Using PV = nRT
<em>Where P is pressure (735torr / 760 = 0.967atm); n are moles (1.394mol); R is gas constant (0.082atmL/molK); T is temperature (28.2°C + 273.15 = 301.35K). </em>
Thus, volume is:
V = nRT / P
V = 1.394mol×0.082atmL/molK×301.35K / 0.967atm
V = 35.6L
<em>The volume of nitrogen oxide formed is 35.6L</em>