Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:
If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:
Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.
Regards.
miscible cause the clear blue liquid is a homogeneous mixture
I think it’s heat transfer but can’t guarantee
Answer:
0.0008 mole of HCl
Explanation:
From the balanced equation between HCl and NaOH:
1 mole of acid requires 1 mole of base for neutralization.
mole = molarity x volume
mole of NaOH = 0.02 x 0.04 = 0.0008 mole
mole HCl = 0.16 x 0.01 = 0.0016 mole
Since 1 mole of NaOH requires 1 mole of HCl for neutralization, it means that 0.0008 mole of NaOH will require 0.0008 mole of HCl. The remaining mole of HCl must have been neutralized by the antiacid:
0.0016 - 0.0008 = 0.0008 mole.
Therefore, 0.0008 mole of HCl must have been neutralized by the antiacid.
So if we use the equation:
→ 
We can then determine the amount of 
needed to produce 208 kg of methanol.
So let's find out how many moles of methanol 208 kg is:
Methanol molar weight = 32.041g/mol
So then we can solve for moles of methanol:
So now that we have the amount of moles produced, we can use the molar ratio (from the balanced equation) of hydrogen and methanol. This ratio is 2:1 hydrogen:methanol.
Therefore, we can set up a proportion to solve for the moles of hydrogen needed:
So now that we have the number of moles of 
that are produced, we can then use the molar weight of hydrogen to solve for the mass that is needed:
Therefore, the amount of diatomic hydrogen () that is needed to produce 208kg of methanol is 
g.
