Answer: 51.45 grams of excess reagent is left after the completion of reaction.
Explanation: For the calculation of moles, we use the formula:
....(1)
Given mass = 92 grams
Molar mass = 28g/mol
Putting values in equation 1, we get:
- For
Given mass = 112 grams
Molar mass = 116g/mol
Putting values in equation 1, we get:
The reaction follows:
By Stoichiometry,
2 moles of reacts with 3 moles of silicon
So, 0.965 moles of reacts with = 
= 1.4475 moles of Silicon.
As, the moles of silicon is more than the required amount and is present in excess.
So, the excess reagent for the reaction is Silicon.
Moles of silicon remained after reaction = 3.285 - 1.4475 = 1.8375 moles
To calculate the amount of Silicon left in excess is calculated by using equation 1:
Amount of Silicon in excess will be 51.45 grams.
Answer:
A. Mixture
Explanation:
Our air has a group of gases. For example, you said nitrogen & oxygen, Which is significantly a mixture.
Answer:
One mole of a substance is equal to 6.022 × 10²³ units of that substance (such as atoms, molecules, or ions). The number 6.022 × 10²³ is known as Avogadro's number or Avogadro's constant. The concept of the mole can be used to convert between mass and number of particles
Answer:
NA = 6.8 E-12 Kg H2(g) / hour
Explanation:
steady-state diffusion of A through non-diffuser B:
- NA = (DAB/RTz)(p*A1 - p*A2)
∴ (A): H2(g)
∴ (B): Pd
∴ DAB = 1.7 E-8 m²/s
∴ p*A1 = 2.0 Kg H2 / m³ Pd
∴ p*A2 = 0.4 Kg H2 / m³ Pd
∴ z = 6 mm = 6 E-3 m
∴ T = 600°C ≅ 873 K
∴ R = 8.314 J/mol.K = 8.314 N.m/mol.K
⇒ NA = ((1.7 E-8)/(8.314)(873)(6 E-3))(2.0 - 0.4)
⇒ NA = 6.246 E-10 mol/s.m³
for A = 0.25 m²
⇒ volume (v) = A×z = (0.25)(6 E-3) = 1.5 E-3 m³
∴ Mw H2(g) = 2.016 g/mol
⇒ NA = (6.246 E-10 mol/s.m³)(1.5 E-3 m³)(2.016 g/mol)(Kg/1000 g)(3600 s/h)
⇒ NA = 6.8 E-12 Kg H2(g)/h
