Question

Question 1Propane gas is commonly used in household grills, barbecues, and camping stoves. Heat for these stoves is produced through thecombustion of propane gas (C3Hg) in the presence of Oz which produces carbon dioxide (CO2) and water (H2O). The density of thispressurized propane gas is usually around 0.52 kilograms of propane per liter. If you use 0.2 L of propane gas during a cookout,how much carbon dioxide in grams does this reaction produce? (Assume 02 is in excess)Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer.a416 gb 3128C2088d 104 gUnanswereda Savemartian

Answer 1

`312\text{ g}`

Step-by-step explanation:

Here, we want to get the mass of carbon (iv) oxide produced

We start by writing the equation of reaction as follows:

`C_3H_(8(g))\text{+5 O}_(2(g))\text{ }\rightarrow\text{ 3CO}_(2(g))\text{ + 4H}_2O_((g))`

Now, we need to get the mass of propane that reacted

We can get that by multiplying the density of propane by its given volume

Mathematically, we have that as:

`\begin{gathered} 0.52\text{ }*0.2\text{ = 0.104 kg} \\ 1000\text{ g = 1kg} \\ 0.104\text{ kg = 0.104 }*\text{ 1000g = 104 g} \end{gathered}`

From here, we get the actual number of moles of propane that reacted

We can get that by dividing the mass by the molar mass of propane

The molar mass of propane is 44 g/mol

The number of moles is thus:

`(104)/(44)\text{ mol}`

From the balanced equation:

1 mole of propane gave 3 moles of carbon (iv) oxide

104/44 mol will give x moles

We have the value of x as:

`x\text{ = }(104)/(44)*\text{ 3 = }(312)/(44)\text{ mol}`

To get the mass of carbon (iv) oxide produced, we multiply the number of moles above by the molar mass of carbon (iv) oxide

The molar mass of carbon (iv) oxide is 44 g/mol

Thus, we have the mass as:

`(312)/(44)*44\text{ = 312 g}`