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Question
112 cm3 of H2S(g) is mixed with 120 cm3 of Cl2(g) at STP to produce HCl(g) and sulphur(s). Calculate composition of the resulting mixture.
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Solution
The composition of resulting mixture is 224cm3 HCl + 8cm3 Cl2
RELATED QUESTIONS
Propane burns in air according to the following equation:
C3H8 + 5O2 → 3CO2 + 4H2O
What volume of propane is consumed on using 1000 cm3 of air, considering only 20% of air contains oxygen?
Calcium carbide is used for the artificial ripening of fruits. Actually the fruit ripens because of the heat evolved while calcium carbide reacts with the moisture. During this reaction calcium hydroxide and acetylene gas are formed. If 200 cm3 of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below.
\[\ce{2C2H2_{(g)} + 5O2_{(g)}-> 4CO2_{(g)} + 2H2O_{(g)}}\]
Propane burns in air according to the following equation:
\[\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}\]
What volume of propane is consumed on using 1000 cm3 of air, considering only 20% of air contains oxygen?
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What volume of air (containing 20% O2 by volume) will be required to burn completely 10 cm3 of methane and acetylene?
\[\ce{CH4 + 2O2 → CO2 + 2H2O}\]
\[\ce{2C2H2 + 5O2 -> 4CO2 + 2H2O}\]
LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to atmosphere.
\[\ce{C3H8 + 5O2 → 3CO2 + 4H2O}\]
\[\ce{2C4H10 + 13O2 → 8CO2 + 10H2O}\]
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- If this ratio of gas volume refers to the reactants and products of a reaction, which gas law is being observed?
- If the volume of A is actually 5.6 dm3 at STP, calculate the number of molecules in the actual Volume of D at STP (Avogadro's number is 6 × 1023).
- Using your answer from (iv), state the mass of D if the gas is dinitrogen oxide (N2O).
112 cm3 of H2S(g) is mixed with 120 cm3 of Cl2(g) at STP to produce HCl(g) and sulphur(s). Calculate the volume of gaseous product formed.
1250 cc of oxygen was burnt with 300cc of ethane [C2H6]. Calculate the volume of unused oxygen formed:
\[\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}\]
