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.
Explanation:
Atomic number (Z) = Number of the protons
It means that if the atomic number of C is 6, it implies that the number of protons in carbon is 6 too.
Mass number (M) = Number of the protons + Number of the neutrons
It means that if the atomic mass of C is 12, it implies that the number of protons and the neutrons in carbon is 12.
Calculating for the moles of H+
1.0 L x (1.00 mole / 1 L ) = 1 mole H+
From the given balanced equation, we can use the stoichiometric ratio to solve for the moles of PbCO3:
1 mole H+ x (1 mole PbCO3 / 2 moles H+) = 0.5 moles PbCO3
Converting the moles of PbCO3 to grams using the molecular weight of PbCO3
0.5 moles PbCO3 x (267 g PbCO3 / 1 mole PbCO3) = 84.5 g PbCO3
Answer:
The answer is b. The number of collisions of gas particles increases