How many grams of o2(g) are needed to completely burn 48.5 g of c3h8(g)?

176g

Explanation:

To appropriately answer this question, we will need to write a complete chemical equation . Firstly what do we know? We know that the molecule given in the question is a hydrocarbon and it burns in oxygen to form carbon iv oxide and water. The reaction equation is written below:

C3H8(g)+5O2(g) --> 3CO2(g)+4H2O(g)

From the equation, we can see that one mole of propane needed 5 moles of oxygen for complete combustion. This is the theoretical relation. Let us find the actual number of moles of propane that burned.

To calculate this, what we simply do is to disburse the mass of propane by the molar mass of propane. The molar mass of propane here is 3(12) + 8(1) = 36 + 8 = 44g/mol

The number of moles of propane is thus = 48.5/44 = 1.102 moles.

Now since one mole of propane needed 5 moles of oxygen , this means 1.102 moles of propane will need 5 * 1.102 moles of oxygen = 5.5 moles

We then proceed to calculate the mass of oxygen gas needed.

The mass of oxygen gas needed equals the number of moles multiplied by the molecular mass of the oxygen molecule. The molecular mass of the oxygen molecule is 2(16) = 32g/mol

The mass needed is thus 5.5 * 32 = 176g

answer; /// the amount of material dissolved in a measure of liquid(concentration) the above statement is (true); /// good luck

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"