Answer:
54.4 mol
Explanation:
the equation for complete combustion of butane is
2C₄H₁₀ + 13O₂ ---> 8CO₂ + 10H₂O
molar ratio of butane to CO₂ is 2:8
this means that for every 2 mol of butane that reacts with excess oxygen, 8 mol of CO₂ is produced
when 2 mol of C₄H₁₀ reacts - 8 mol of CO₂ is produced
therefore when 13.6 mol of C₄H₁₀ reacts - 8/2 x 13.6 mol = 54.4 mol of CO₂ is produced
therefore 54.4 mol of CO₂ is produced
Answer:
46.3g H2O
Explanation:
start by balancing it: CaC2(s) + 2H2O(g) -> Ca(OH)2(s) + C2H2(g)
then use factor label method to solve
82.4g CaC2 x (1 mol CaC2/64.10g CaC2) x (2 mol H2O/1 mol CaC2) x (18.016g H2O/1 mol H20) = 46.3g H2O
Answer:
One mole of water was produced from this reaction.
Explanation:
According to this question, the following equation is given as follows:
2H2 + O2 → 2H2O
Two (2) moles of hydrogen gas produces two (2) moles of water in this balanced chemical equation.
If 1 mole of hydrogen gas was used, then:
1 × 2/2 moles of water would be produced
1 mole of water would be produced.
False, kinetic energy depends on an objects mass and speed
