the ideal gas law pvnrt relates pressure p volume v temperature t and number of moles of a gas n

1. First, we need to find the number of moles of CO2 produced from 41.0 L of CO2 at STP (Standar

The ideal gas law pvnrt relates pressure p volume v...

The ideal gas law, PV = nRT, relates pressure (P), volume (V), temperature (T), and the number of moles of a gas (n). The gas constant (R) equals 0.08206 L·atm/(K·mol) or 8.3145 J/(K·mol). The equation can be rearranged as follows to solve for n: n = PV/RT. This equation is useful when dealing with gaseous reactions because stoichiometric calculations involve mole ratios. CaCO3(s) → CaO(s) + CO2(g) 1. How many grams of calcium carbonate are needed to produce 41.0 L of carbon dioxide at STP? 2. Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation for the complete combustion of butane is: 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l) At 1.00 atm and 23 °C, how many liters of carbon dioxide are formed by the combustion of 1.40 g of butane?

Transcript

00:01 Hi, so we have two questions here.

00:04 First one, we're solving for the mass of caco3 or calcium carbonate required to produce the given volume of co2 carbon dioxide at stp.

00:15 So for the first reaction, we have calcium carbonate forming calcium oxide and co2.

00:23 So this is the balance equation for this.

00:25 First, we will use the ideal gas equation to solve for the number of moles of co2.

00:30 So ideal gas equation is pv is equivalent to nrt.

00:35 We are arranging this to solve for n, the number of moles of co2 will be equivalent to pv over rt.

00:41 Now let's plug in the information that we have.

00:44 We have here stp and remember that stp stands for standard temperature and pressure.

00:50 And the temperature is equivalent to 0 degrees celsius or 273 .15 kelvin.

00:54 And pressure is 1 atm.

00:57 Plug in the information that we have.

00:59 Pressure is 1 atm.

01:01 Volume is 41 liters.

01:06 R is 0 .08206 liters atm per mole kelvin.

01:13 Temperature is 273 .15 kelvin and we'll cancel some units now.

01:18 We could cancel atm, liters, and moles.

01:22 And as you can see, the remaining unit is moles.

01:26 Let's use other color.

01:32 Now solving for this one, the number of moles of co2 is therefore equivalent to.

01:38 This is 1 .829 moles of carbon dioxide produced.

01:43 Now we will use the chemistry to solve for the mass of the reactant calcium carbonate.

01:51 If we are to produce 1 .829 moles of co2, we'll multiply this stoichiometric ratio here as per the given balance equation.

01:59 For every 1 mole of co2 produced, it will require 1 mole of the reactant caco3.

02:06 Multiply the molar mass of this compound.

02:10 That will be 100 .09 grams per mole of calcium carbonate.

02:17 So we can now cancel some units, moles of co2, moles of caco3...

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