The balanced chemical reaction is:
<span>C6H12 + 9O2 = 6CO2 + 6H2O
We are given the amount of </span><span>C6H12 to be used for the reaction. THis will be the starting point for the calculations.
1.900 g </span>C6H12 ( 1 mol C6H12/ 84.18 g C6H12) ( 6 mol CO2 / 1 mol <span>C6H12) (44.01 g /mol ) = 5.96 g CO2
</span>1.900 g C6H12 ( 1 mol C6H12/ 84.18 g C6H12) ( 6 mol H2O / 1 mol <span>C6H12) (18.02 g /mol ) = 2.44 g H2O
</span>
Answer:
6.2g of NaBr are produced
Explanation:
The reaction of HBr with NaOH occurs as follows:
HBr + NaOH → NaBr + H2O
<em>Where 1 mole of each reactant produce 1 mole of NaBr</em>
To solve this question we need to find the moles of each reactant using their molar mass. With moles we can find limiting reactant and the moles (And mass) of NaBr produced, as follows:
<em>Moles HBr -Molar mass: 80.9119g/mol)-</em>
4.9g * (1mol/80.9119g) = 0.0606 moles HBr
<em>Moles NaOH -Molar mass: 40g/mol-</em>
3.86g * (1mol/40g) = 0.0965 moles NaOH
As the reaction is 1:1 and the moles of HBr < Moles NaOH, the limiting reactant is HBr and moles of NaBr produced are 0.0606 moles.
The mass of NaBr (Molar mass: 102.894g/mol) is:
0.0606 moles * (102.894g/mol) =
<h3>6.2g of NaBr are produced</h3>
Answer:
No
Explanation:
You can not have a reaction that produces "no mass". This breaks the law of conservation of mass. When gasses are produced, they will have a specific amount of mass associated with them which means they will have a weight. They will have to be accounted for