chemistry help please?

Question

Nitrogen dioxide has opposing roles in our world. It is essential in the industrial production of nitric acid, but is also a contributor to acid rain and photochemical smog. Calculate the volume of nitrogen dioxide produced at 741 torr and 28.2°C by the reaction of 4.97 cm3 of copper (d = 8.95 g/cm3) with 230.0 mL of nitric acid (d = 1.42 g/cm3, 68.0% HNO3 by mass).

Cu(s) + 4 HNO3(aq) ===> Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l)

Answer 1

First we need to know which reagent is in excess and which is not.

First the mass of copper is calculated, which is the volume of copper times its density= 4.97 *8.95= 44.4815g. This must be converted into moles by dividing the mass by its molar mass of copper(=63.55g.mol)= 44.4815/63.55= 0.70moles.

Next the mass of nitric acid needs to be calculated, which is the volume of nitric acid multiplied by its density multiplied by its composition= 230*1.42*0.68= 222.088. This must be converted into moles by dividing the mass by its molar mass of nitric acid(=63.01.mol)= 222.088/63.01= 3.5246 moles.

From this we know that Copper is total consumed and nitric acid is in excess. From the above equation we know that 1 mole of copper will produce 2 moles of nitrogen dioxide, therefore 1.4 moles of NO2 will be produced.

We can use the ideal gas law to work out the pressure of the gas.

PV=nRT, or V=nRT/P

where V is volume, P is Pressure, n is the number of moles, R is the Gas Constant, T is temperature

V= 1.4 *8.314* (301.35)/(133.3223684*741)= 0.0355 cubic metres = 35.50 litres.

Answer 2

First you need to find the limiting reactant, thus you need to find the moles of each reactant.

4.97 cm^3 * 8.95 g/cm^3 = 44.5 g / 63.5 g/mole = 0.701 moles Cu

230 mL * 1.42 g/mL = 327 g HNO3 solution * 0.680 = 222 g HNO3 / 63.0 g/mol = 3.52 moles HNO3

Cu is your limiting reactant, so by the equation, you will get 2*0.701 or 1.402 moles of NO2

Plug this into the Ideal Gas Law to get the volume (convert pressure to atm and temperature to K).

PV = nRT
V = nRT/P = 1.40 * 0.08206 * 301.4/0.975 = 35.5 L

Answer 3

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