Stoichiometry And Stoichiometric Calculations

Balanced equation: C H + 5O → 3CO + 4H O
From 5 moles of O , 4 moles of H O form. So, 2.5 moles of O will form:

To determine the moles of OH , we use the stoichiometry of the reaction, where 1 mole of CrO reacts with 3 moles of SnO to form OH .

The balanced chemical equation for the combustion of carbon monoxide is: From the equation, 2 moles of CO react with 1 mole of O to produce 2 moles of CO .
The molar mass of CO is 28 g/mol, and the molar mass of CO is 44 g/mol.
To calculate the volume of oxygen needed to produce 22 g of CO , we first find the number of moles of CO : Since 1 mole of O reacts with 2 moles of CO to produce 2 moles of CO , the moles of O required are: At STP, 1 mole of any ideal gas occupies 22.4 L, so the volume of O is:
Thus, the correct answer is (b).

The reaction for the burning of calcium carbonate is: From the reaction, we know the molar mass of calcium carbonate (CaCO ) is 100 g/mol, and calcium oxide (CaO) is 56 g/mol.
The stoichiometric ratio is 1:1, meaning that 100 g of calcium carbonate gives 56 g of calcium oxide.
Therefore, if 28 g of calcium oxide is produced, the mass of calcium carbonate required is:
Thus, the correct answer is (d).

Given that the molecular weight of X is 148 g/mol and the molecular weight of Be is 9 g/mol, we can calculate the molar masses of N, Cl, and H from the given percentages. The total moles of each element in the compound can be found using the following relationships:

- The moles of Be =
- The moles of N =
- The moles of Cl =
- The moles of H =

By calculating the number of atoms of each element, we can determine that the compound’s formula is BeN4Cl2H12. This formula is consistent with the molar masses and the percentages of each element in the compound.

Therefore, the correct molecular formula of the compound is (a) BeN4Cl2H12.