3 Kg mass of water is heated to a temperature of 99.5 C at 1 bar of pressure. It is slowly poured into a heavily insulated beaker containing 7.5 Kg of water at a temperature and pressure of 4 C, 1 bar, respectively. The specific heat of the water, an incompressible material is, c = 4.1 KJ/(Kg K). The beaker can be considered to be an adiabatic system. The two water mass’ reach an equilibrium temperature. There is no kinetic...
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Richard G.
Chemistry 101
2 weeks, 6 days ago
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When methane (CH4) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reaction is CH4(g)+O2(g)→CO2(g)+H2O(g) a).What mass of carbon dioxide is produced from the complete combustion of 1.90×10−3 g of methane? b)What mass of water is produced from the complete combustion of 1.90×10−3 g of methane? c) What mass of oxygen is needed for the complete combustion of 1.90×10−3 g of methane?
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Chris E. When methane (CH4) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reaction is CH4(g)+O2(g)→CO2(g)+H2O(g)This type of reaction is referred to as a complete combustion reaction
1 Expert Answer
Nathan I. answered • 04/23/19
Patient, Knowledgeable Tutor in Chemistry
First, we need to balance the stoichiometry equation.
?CH4(g) + ?O2(g) --> ?CO2(g) + ?H2O(g)
Lets start by putting 1 for each question mark. There are 2 hydrogens in water, and 4 in methane. This means whatever number we put in front of methane, the number in front of H2O has to be double that. If we put a 2 in front of H2O, the only thing unbalanced in the equation in the number of oxygens: there are four on the right side and only two in the oxygen on the left. We can make the coefficient in front of oxygen 2:
1CH4(g) + 2O2(g) --> 1CO2(g) + 2H2O(g)
But, to use the stoichiometry equation, we need to figure out the number of moles of methane used, instead of grams. The molecular weight of methane is 16.043g/mol. We can use this to find the moles of methane:
1.2*10^-3g / (16.043g/mol)=7.48*10^-5 moles of methane
Now that we know the number of moles of methane combusted, we can use the stoichiometry equation to find out how many moles of CO2 this equates to. For every mole of methane burned, a mole of CO2 is produced:
7.48*10^-5 mol CH4 * (1 mol CO2 / 1 mol CH4) = 7.48*10^-5 mol CO2
We know that the reaction goes to completion, so all the methane will be reacted. To find the mass of carbon dioxide used, we can use the number of moles from the above equation, and the molecular weight of CO2 (= 44.009 g/mol).
7.48*10^-5 mol CO2 * 44.009 g/mol = 3.29*10^-3 g CO2
The mass of CO2 produced is 3.29*10^-3g.
We first need to write down the balanced equation for this reaction, that is
#CH_4(g)+2O_2(g)->CO_2(g)+2H_2O(g)#
As we can see here, the mole ratio between methane #(CH_4)# and carbon dioxide #(CO_2)# is #1:1#, i.e. one mole of methane will produce one mole of carbon dioxide in return.
In here, we burn #2.6*10^-3 \ "g"# of methane. So, we need to convert that amount into moles.
Methane has a molar mass of #16 \ "g/mol"#. So, in here, we have
#(2.6*10^-3cancel"g")/(16cancel"g""/mol")=0.0001625~~1.63*10^-4 \ "mol of" \ CH_4#
Since the mole ratio between the two compounds is #1:1#, then we will also produce #1.63*10^-4# moles of carbon dioxide.
Carbon dioxide has a molar mass of #44 \ "g/mol"#.
So, the mass of carbon dioxide produced is
#1.63*10^-4cancel"mol"*44 \ "g/"cancel"mol"=0.007172 \ "g"#
Now, we can convert into scientific notation, which will be
#7.17*10^-3 \ "g"#
Since this answer asks us to give the answer in appropriate units, I suggest you use milligrams as the answer, as #1 \ "g"=1*10^3 \ "mg"#.
#<=>7.17*10^-3 \ "g" = 7.17 \ "mg"#