Answer:
This question appear incomplete
Explanation:
This question appear incomplete because an equation to show the production of sodium chloride from calcium chloride should have been illustrated. However, if the balanced chemical equation showing sodium chloride (NaCl) been a product of a reaction involving calcium chloride (CaCl₂) as a reactant (shown below) is to be used, then we start by writing a complete balanced chemical equation
CaCl₂ + Na₂CO₃ ⇒ CaCO₃ + 2NaCl
From the equation above, it can be deduced that 1 mole of CaCl₂ is required to produce 2 moles of NaCl, thus how many moles of CaCl₂ will be required to produce 6.5 moles of NaCl.
1 mole of CaCl₂ = 2 moles of NaCl
? moles of CaCl₂ = 6.5 moles of NaCl
cross multiply
? moles of CaCl₂ = 6.5 × 1/2
? moles of CaCl₂ = 3.25 moles of CaCl₂
3.25 moles of CaCl₂ will be needed to produce 6.5 moles of NaCl
Answer: SnO2 + 2 H2 = Sn + 2 H2O
Explanation: I used a balance equation website. It's called WebQC if you want to check it out for future help.
The balanced chemical reaction is:
CH4 + 2O2 —> CO2 + 2H2O
You need to convert mass to moles (divide by molar mass):
CH4 moles = 5 / 16 = 0.31 mol
O2 moles = 5 / 32 = 0.16 mol
To figure out which reactant is limiting, divide the actual moles by the corresponding coefficient in the reaction:
CH4: 0.31 / 1 = 0.31
O2: 0.16 / 2 = 0.08
O2 is the lower number, so it is the limiting reactant. From the reaction we know it takes 2 moles of O2 to react with each mole of CH4. Therefore, for however many moles of O2 we actually have, half as many moles of CH4 will react. Since we have 0.16 mol of O2, only 0.08 mol of CH4 will react, leaving behind 0.31 - 0.08 = 0.23 mol of CH4.
Now convert back to mass (multiply by molar mass) to find the mass of CH4 remaining:
0.23 x 16 = 3.68g
The closest answer is B.
Answer: Kc= 4.653
Explanation:
From The equation of reaction
CO(g)+2H2(g)⇌CH3OH(g)
From the question
[CH3OH] =0.0679 moles
[CO]= 0.160moles
[H2]= 0.302
Kc = [CH3OH]/[CO] [H2]^2
Kc=[0.069]/[0.16]×[0.302]^2
Kc= 4.653
