2. What is the volume of oxygen required to burn completely 50cm3 of ethyne(C2H2).
3. In the reaction between sodium chloride and sulphuric acid. Calculate the volume of HCL gas that can be obtained at stp from 5.85g of sodium chloride.( H=1,Na= 23,Cl=35.5. molar volume of gas at stp is 22.4dm3)
2. The balanced equation for the combustion of ethyne is:
2C2H2 + 5O2 → 4CO2 + 2H2O
From the equation, we can see that for every 2 moles of C2H2, 5 moles of O2 are required. Therefore, to burn completely 1 mole of C2H2, we would need 5/2 moles of O2.
Knowing that the molar volume of gas at STP is 22.4 dm3, we can calculate the volume of O2 required to burn completely 1 mole of C2H2:
(5/2) x 22.4 dm3 = 56 dm3
Therefore, to burn completely 50 cm3 of C2H2, we would need:
(56 dm3 / 1 mole) x (50 cm3 / 22.4 dm3) = 125 cm3 of O2.
So, the volume of oxygen required to burn completely 50 cm3 of ethyne is 125 cm3.
3. The balanced equation for the reaction between sodium chloride (NaCl) and sulfuric acid (H2SO4) is:
NaCl + H2SO4 → NaHSO4 + HCl
We can see from the equation that for every 1 mole of NaCl, 1 mole of HCl gas is produced.
First, we need to calculate the number of moles of NaCl in 5.85 g:
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Number of moles of NaCl = mass / molar mass = 5.85 g / 58.5 g/mol = 0.1 mol
Since 1 mole of NaCl produces 1 mole of HCl gas, we can say that 0.1 mol of NaCl will produce 0.1 mol of HCl gas.
Finally, we can calculate the volume of HCl gas at STP:
Volume of HCl gas = number of moles x molar volume at STP
= (0.1 mol) x (22.4 dm3/mol)
= 2.24 dm3
So, the volume of HCl gas that can be obtained at STP from 5.85 g of sodium chloride is 2.24 dm3.