2018年3月7日 · C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O We have: C_3H_8 + O_2 -> CO_2 + H_2O I would always start by balancing the elements that only occur once on each side. In this case it's carbon. We also notice that as it now stands we have an odd number of oxygen molecules on the right and an even number on the left, which won't work. We can fix this by multiplying the water by an even number. C_3H_8 + O_2 -> CO ...

C3H8 + 02 ----> 3CO2 + H20 Now the carbons are balanced, we will look at the hydrogens. We can multiply the number of waters on the product side by 4, to make 4 H20 molecules. This gives both sides of the equation 8 hydrogens. Next, we can check our oxygens. There will be 10 oxygens on the products side and 2 on the reactants.

2015年11月19日 · C_3 H_8 + 5O_2 ->3CO_2 +4H_2 O This equation represents the combustion of proprane (a 3 carbon alkane) and it it highly exothermic, gives off heat, delta H < 0.

2015年11月24日 · I would balance it stoichiometrically. C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(g) Is mass balanced here? How do you know?

2015年12月16日 · How would you balance the following equation: C3H8 + O2 --> CO3 + H20?

C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(l) + Delta And here garbage in equals garbage out. This process of balancing, of stoichiometry may seem a little unfamiliar...but only in the context of chemical equations...MASS is CONSERVED...and the usual rigmarole is to balance the carbons as carbon dioxide, balance the hydrogens as water, and THEN balance the oxygens. In a more familiar context ...

2014年6月4日 · Two molecules of "C"_3"H"_8 are required. > The balanced equation is "C"_3"H"_8 + "5O"_2 → "3CO"_2 + "4H"_2"O" This tells you that five molecules of "O"_2 require one molecule of "C"_3"H"_8, so Ten molecules of "O"_2 require twice as many molecules of "C"_3"H"_8 or two molecules. You might be able to do this problem in your head. But it isn't as easy when you have more complicated numbers ...

7.72 "g H"_2"O" The NIE for the complete combustion of propane is: "C"_3 "H"_8 + 5 "O"_2 -> 3 "CO"_2 + 4 "H"_2"O" The problem doesn't state how much propane is burned, so we're going to assume that there is enough to use up all the oxygen. To find out how much that is, we need to convert liters of "O"_2 to moles of "O"_2 using the fact …

So we need more on the right hand side; we already have 2 hydrogen atoms, so to get to 8, we can put a 4 in front of the H2O as 4 x 2 =8. The equation is now C3H8 + O2 —> CO2 + 4H2O Now, we can look at the carbons; on the left hand side we have 3 carbons whereas on the right hand side one, so more carbon atoms are required on the right hand side.

2018年4月25日 · 95.3g C_3H_8 First, check to ensure that the equation is balanced (it is). Then calculate the required moles of product water for a mass of 156 grams. Then use the ratio of moles of product H_2O to reactant C_3H_8 to find the moles of propane required. Finally, use that value with the molecular weight of propane (44) to find the mass of propane required. 156/18 = 8.67 moles H_2O (1 "mol" C_3H ...