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.
Answer:
m = 190. g MgCl2
Explanation:
Use a molar ratio to get the # of moles MgCl2 produced from 4.00 mol of HCl:
4.00 mol HCl × (1 mol MgCl2/2 mol HCl) = 2.00 mil MgCl2
So 2.00 moles of MgCl2 will be produced. To find the mass in grams, use the molar mass of MgCl2:
2.00 mol MgCl2 × (95.211 g MgCl2/1 mol MgCl2)
= 190. g MgCl2
Mass percentage of sodium chloride(NaCl) in ocean waters = 3.5 %
That means 3.5 g sodium chloride(NaCl) is present for every 100 g of ocean water.
The given mass of sodium chloride(NaCl) is 45.8 g
Calculating the mass of ocean waters that would contain 45.8 g sodium chloride(NaCl):
= 1309 g ocean water
Therefore, 45.8 g sodium chloride is present in 1309 g ocean water.
Multiply,
moles * molar mass = mass
