Answer: 10.2 grams
Explanation:
The balanced chemical reaction is :
According to the ideal gas equation:
P = Pressure of the gas = 740 torr = 0.97 atm (760torr=1atm)
V= Volume of the gas = 12.0 L
T= Temperature of the gas = 19°C = 292 K 
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas
According to stoichiometry:
2 moles of hydrogen are generated by = 1 mole of 
Thus 0.48 moles of hydrogen are generated by =
moles of
Mass of 
Thus 10.2 grams of are needed to generate 12.0 L of hydrogen gas if the pressure of hydrogen is 740. torr at 19°C
First, we apply the law of conservation of mass which states that the total mass in a system remains constant.
Therefore, there must be 5.00 g of sulfur and 4.99 g of oxygen in the product. Now, we determine the mass percentage using:
Mass % = (mass of sulfur x 100) / total mass of compound
Mass % = (5 * 100) / (5 + 4.99)
Mass % = 50.05%
The product contains 50.05% sulfur by mass.
Answer:
16.9g of H₂O can be formed
Explanation:
Based on the chemical reaction, 2 moles of H₂ react per mole of O₂. To anser this question we must find limiting reactant converting the mass and volume of each reactant to moles:
<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>
8.76g * (1mol / 2.016g) = 4.345 moles
<em>Moles O₂:</em>
PV = nRT
PV/RT = n
P = 1atm at STP
V = 10.5L
R = 0.082atmL/molK
T = 273.15K at STP
n = 1atm*10.5L / 0.082atmL/molK*273.15K
n = 0.469 moles of oxygen
For a complete reaction of 4.345 moles moles of hydrogen are required:
4.345 moles H2 * (1mol O2 / 2mol H2) = 2.173 moles of O2 are required. As there are just 0.469 moles, Oxygen is limiting reactant
Now, 1 mole of O2 produce 2 moles of H2O. 0.469 moles will produce:
0.469 moles O₂ * (2 moles H₂O / 1mol O₂) = 0.938 moles H₂O.
The mass is -Molar mas H₂O = 18.01g/mol-:
0.938 moles * (18.01g/mol) =
<h3>16.9g of H₂O can be formed</h3>
