The volume of oxygen at STP required would be 252.0 mL.
<h3>Stoichiometic problem</h3>
The equation for the complete combustion of C2H2 is as below:
The mole ratio of C2H2 to O2 is 2:5.
1 mole of a gas at STP is 22.4 L.
At STP, 100.50 mL of C2H2 will be:
100.50 x 1/22400 = 0.0045 mole
Equivalent mole of O2 according to the balanced equation = 5/2 x 0.0045 = 0.01125 moles
0.01125 moles of O2 at STP = 0.01125 x 22400 = 252.0 mL
Thus, 252.0 mL of O2 gas will be required at STP.
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NH3 has a Molar mass of 17g
3.47 * M(NH3) = 58.99g of NH3 in the cleaning agent
Answer:
9.8g
Explanation:
Using periodic table find molar mass of C and H:
C=12.01g/mol
H=1.008g/mol
Molar mass of C2H4=(12.01)2+(1.008)4=28.03g/mol
Using molar mass times moles of the chemical to find the mass in 0.35 moles of c2h4:
28.03 x 0.35=9.8grams
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Answer: chlorine gas
Explanation:
mass = 0.508 g volume = 0.175 L
From ideal gas equation.
PV = nRT
(1.0 atm) × (0.175 L) = n × [0.0821 atm L / (mol K)] × [(273.2 + 25.0) K]
n = 1.0 × 0.175 / (0.0821 × 298.2) mol
n = 0.00715 mol
Molarmass= mass/mole
Molar mass of the gas = (0.508 g) / (0.00715 mol) = 71.0 g/mol
The molar mass of Cl₂ (chorine) = 35.5 × 2 g/mol = 71.0 g/mol
Hence, the gas is chlorine.
