50cm/3 ofsulphur(iv) oxide were produced at stp when some quality of powdered sulpur were burnt in excess oxygen. (1) which equation is suitable.(2) what volume of oxygen was used up during the reaction.(3) which of the gas law is applicable. State the law.
You need to clean up the post Is that S2O8?. There are no choices for equations to use.
1) S O2 -> SO2
2) USING IDEAL GAS EQUATION. PV=nRT.... 760V=32R(273).,V=8736/760...V=11.5cm.
3) IDEAL GAS LAW EQUATION.
4) THIS EQUATION STATES THAT FOUR QUANTITIES ARE IMPORTANT IN ALL EXPERIMENTAL WORK,MEARSUREMENT OR CALCULATION INVOLVING GASES. . .THEY ARE Volume,Pressure,Temperature,Number of moles
Pls promise where is the valuefor
r in that question
To answer these questions, we need to follow a step-by-step approach. Let's break it down:
(1) Finding the Suitable Equation:
To determine the suitable equation, we need to calculate the number of moles of sulfur dioxide produced. We can do this by using the relationship between moles, mass, and molar mass.
Given:
- 50 cm³ of sulfur dioxide (SO₂) produced.
- We have an excess of oxygen.
Step 1: Convert the volume of SO₂ to moles.
To convert the volume to moles, we need to use the molar volume of a gas at STP, which is 22.4 L/mol.
1 L = 1000 cm³
So, 22.4 L/mol = 22,400 cm³/mol.
50 cm³ × (1 mol / 22,400 cm³) ≈ 0.00223 moles of SO₂
Step 2: Write out the balanced chemical equation.
We know that sulfur is burned in excess oxygen to produce sulfur dioxide.
Sulfur (S) + Oxygen (O₂) → Sulfur Dioxide (SO₂)
Since the question specifies "Sulphur(IV) oxide", which is sulfur dioxide (SO₂), we have the correct equation.
Answer: The suitable equation for this reaction is S + O₂ → SO₂.
(2) Calculating the Volume of Oxygen Used:
To find the volume of oxygen used, we need to use the stoichiometry of the balanced chemical equation.
From the equation: S + O₂ → SO₂
1 mole of sulfur reacts with 1 mole of oxygen to produce 1 mole of sulfur dioxide.
Given 0.00223 moles of sulfur dioxide were produced, we can conclude that the same number of moles of oxygen were used.
Therefore, the volume of oxygen used can be calculated using the molar volume of a gas at STP:
Volume of oxygen = moles of oxygen × molar volume at STP
Volume of oxygen = 0.00223 moles × 22.4 L/mol ≈ 0.05 L or 50 cm³
Answer: The volume of oxygen used up during the reaction is approximately 50 cm³.
(3) Applicable Gas Law:
To determine which gas law is applicable, let's consider the conditions of the given problem:
- The reaction is at STP (Standard Temperature and Pressure), which means a temperature of 273 K (0 °C) and a pressure of 1 atmosphere (atm).
The ideal gas law (PV = nRT) should be applicable here. This law states that the pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas are related. It can be expressed as:
PV = nRT
Here:
- P = 1 atm (from STP)
- V = 50 cm³ (volume of oxygen used)
- n = number of moles of oxygen (0.00223 mol, calculated earlier)
- R = gas constant (0.0821 L·atm/(mol·K))
- T = 273 K (from STP)
Answer: The applicable gas law is the ideal gas law (PV = nRT).
So, to summarize:
(1) The suitable equation for the reaction is S + O₂ → SO₂.
(2) The volume of oxygen used up during the reaction is approximately 50 cm³.
(3) The applicable gas law is the ideal gas law (PV = nRT).