Answer:
7.46 g.
Explanation:
- Firstly, we need to calculate the amount of heat needed to warm 5.64 kg of water from 21.0°C to 70.0°C using the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by water (Q = ??? J).
m is the mass of water <u><em>(m: we will determine).</em></u>
c is the specific heat capacity of water (c = 4.186 J/g.°C).
ΔT is the temperature difference (final T - initial T) (ΔT = 70.0 °C - 21.0 °C = 49.0 °C).
- To determine the mass of 1.76 L of water we can use the relation:
mass = density x volume.
density of water = 1000 g/L & V = 1.76 L.
∴ mass = density x volume = (1000 g/L)(1.76 L) = 1760.0 g.
∵ Q = m.c.ΔT
<em>∴ Q = m.c.ΔT </em>= (1760.0 g)(4.186 J/g.°C)(49.0 °C) = 360483.2 J ≅ 360.4832 kJ.
- As mentioned in the problem the molar heat of combustion of hexane is - 4163.0 kJ/mol.
<em>Using cross multiplication we can get the no. of moles of hexane that are needed to be burned to release 360.4832 kJ:</em>
Combustion of 1.0 mole of methane releases → - 4163.0 kJ.
Combustion of ??? mole of methane releases → - 360.4832 kJ.
∴ The no. of moles of hexane that are needed to be burned to release 360.4832 kJ = (- 360.4832 kJ)(1.0 mol)/(- 4163.0 kJ) = 0.0866 mol.
- Now, we can get the mass of hexane that must be burned to warm 1.76 L of water from 21.0°C to 70.0°C:
<em>∴ mass = (no. of moles needed)(molar mass of hexane)</em> = (0.0866 mol)(86.18 g/mol) = <em>7.46 g.</em>
B reactants then C is on the right side! If you take h2O and air (the reactants) then the product would be a snowflake. happy holidays!!
Answer:
5.The limiting reactant is completely used up in the reaction
Explanation:
The limiting reactant is completely used up in the reaction is the correct answer because a limiting reactant is a reactant in chemical reaction that is completely consumed or used up in the chemical reaction. Limiting reactant when it is completely used up limits the amount of products that will be formed. The reaction will be halted or will stop when the limiting reactant is totally used up.
