Answer: 5.72 g mass of carbon dioxide was produced
Explanation:
Moles of C =
Mass of reacted = 59.1 g - 17.5 g = 41.6 g
Moles of
According to reaction 1 mole produces 1 mole of
then 1.3 mole of
will produce
moles of
Mass of :
=Moles of × molar mass of
= 1.3 ×44 g/mol =
=57.2 grams
5.72 g mass of carbon dioxide was produced
Answer:
The mass of hydrogen gas in the mixture: <u>w₂ = 0.433 g</u>
Explanation:
<u>According to the ideal gas equation: </u>
for an ideal gas,
and
Here, P: total pressure of the gases = 2.14 atm
V: total volume of the gases = 6.68 L
T: temperature = 19 °C = 19+273.15 = 292.15K (∵ 0°C = 273.15K)
R: gas constant = 0.08206 L·atm·K⁻¹·mol⁻¹
: total number of moles of gases
<u>To calculate the total number of moles of gases</u>:
= <u>0.5963 moles</u>
Let, the number of moles of carbon dioxide be n₁ and number of moles of hydrogen be n₂
<u>Given:</u> mass of carbon dioxide: w₁ = 16.8 g, mass of hydrogen: w₂ = ?g
molar mass of carbon dioxide: m₁ = 44.01 g/mol, molar mass of hydrogen: m₂= 2.016 g/mol
Therefore, = n_{1}+n_{2} = (w₁ ÷ m₁) + (w₂ ÷ m₂)
⇒ 0.5963 mol = (16.8 g ÷ 44.01 g/mol) + (w₂ ÷ 2.016 g/mol)
⇒ 0.5963 mol = (0.3817mol) + (w₂ ÷ 2.016 g/mol)
⇒ 0.5963 mol - 0.3817mol = (w₂ ÷ 2.016 g/mol)
⇒ 0.2146 mol = (w₂ ÷ 2.016 g/mol)
⇒ w₂ = 0.433 g
<u>Therefore, the mass of hydrogen gas in the mixture: w₂ = 0.433 g</u>