Answer:
a) Limiting: sulfur. Excess: aluminium.
b) 1.56g Al₂S₃.
c) 0.72g Al
Explanation:
Hello,
In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:
a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:
Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.
b) By stoichiometry, the produced grams of aluminium sulfide are:
c) The leftover is computed as follows:
NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.
Best regards.
Answer:
Conociendo el volumen de solución, masa de soluto y su masa molar, es posible determinar: B) Concentración molar
La molaridad es la relación entre el número de moles de soluto y los litros de solución. Más:
M = No moles de solución de soluto / volumen (L)
Y a su vez los moles de soluto se encuentran por:
No moles de soluto = masa soluto / masa molar soluto
Correct me if I'm wrong, anyone who is more informed on this than I am, but wouldn't it be because of the magnetic fields and whatnot? The longer I think on it, the more I feel like I'm wrong, so don't take my word for it 100%
Answer:
because iron is magnetic and sand is not
Explanation:
<u>Answer:</u> The mass of solid NaOH required is 80 g
<u>Explanation:</u>
Equivalent weight is calculated by dividing the molecular weight by n factor. The equation used is:
where,
n = acidity for bases = 1 (For NaOH)
Molar mass of NaOH = 40 g/mol
Putting values in above equation, we get:
Normality is defined as the umber of gram equivalents dissolved per liter of the solution.
Mathematically,
Or,
......(1)
We are given:
Given mass of NaOH = ?
Equivalent mass of NaOH = 40 g/eq
Volume of solution = 400 mL
Normality of solution = 5 eq/L
Putting values in equation 1, we get:
Hence, the mass of solid NaOH required is 80 g
