The combustion of methane (CH4) can be represented by the balanced chemical equation:
CH4 + 2O2 → CO2 + 2H2O.
To find the mass of methane required to produce 22 g of CO2, we proceed as follows:
Step 1: Determine the molar mass of CO2
Molar mass of CO2 = C + 2(O) = 12.0 + 2(16.0) = 44.0 g/mol.
Step 2: Calculate the moles of CO2 produced
Moles of CO2 = mass/molar mass = 22 g / 44.0 g/mol = 0.5 mol.
Step 3: Use the stoichiometry of the reaction to find moles of CH4
According to the balanced equation, 1 mole of CH4 produces 1 mole of CO2. Therefore, 0.5 mol of CO2 requires 0.5 mol of CH4.
Step 4: Determine the mass of CH4
Molar mass of CH4 = C + 4(H) = 12.0 + 4(1.0) = 16.0 g/mol.
Mass of CH4 = moles × molar mass = 0.5 mol × 16.0 g/mol = 8 g.
The mass of methane required is 8 g,
CH4 + 2O2 → CO2 + 2H2O
Moles of CO2 produced:
\[ \frac{22}{44} = 0.5 \, \text{mol} \]
Required moles of CH4:
\[ 0.5 \, \text{mol} \times 16 \, \text{g/mol} = 8 \, \text{g} \]

Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \( f(x + y) = f(x) f(y) \) for all \( x, y \in \mathbb{R} \). If \( f'(0) = 4a \) and \( f \) satisfies \( f''(x) - 3a f'(x) - f(x) = 0 \), where \( a > 0 \), then the area of the region R = {(x, y) | 0 \(\leq\) y \(\leq\) f(ax), 0 \(\leq\) x \(\leq\) 2 is :
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is: