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 …

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 …

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 …

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 …

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 …

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 …

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 …