Answer:
164.3g of NaCl
Explanation:
Based on the chemical equation:
CaCl2 + 2NaOH → 2NaCl + Ca(OH)2
<em>where 1 mole of CaCl2 reacts with 2 moles of NaOH</em>
To solve this question we must convert the mass of CaCl2 to moles. Using the chemical equation we can find the moles of NaCl and its mass:
<em>Moles CaCl2 -Molar mass: 110.98g/mol-</em>
156.0g CaCl₂ * (1mol / 110.98g) = 1.4057 moles CaCl2
<em>Moles NaCl:</em>
1.4057 moles CaCl2 * (2mol NaCl / 1mol CaCl2) = 2.811 moles NaCl
<em>Mass NaCl -Molar mass: 58.44g/mol-</em>
2.811 moles NaCl * (58.44g / mol) = 164.3g of NaCl
Products are copper+ aluminium chloride
reactants are aluminium+copper chloride
Answer:
180g
Explanation:
H:1 O:16
2H2+O2 → 2H2O
2 2(16) 2(1)+(16)
32 18
Now,
32g of O → 2(18)g of H2O
160g of O → 2(18)g divides by 32g times 160g
=180g
The balanced chemical equation for the above reaction is as follows;
2Ca + O₂ --> 2CaO
stoichiometry of Ca to O₂ is 2:1
this means that 2 mol of Ca reacts with 1 mol of O₂.
If O₂ is the limiting reactant,
4 mol of O₂ should react with (4x2) - 8 mol of Ca
however only 7.43 mol of Ca is present. Therefore Ca is the limiting reactant.
7.43 mol of Ca reacts with - 7.43/2 = 3.715 mol of O₂
therefore there's excess O₂₂ remaining after the reaction
Since Ca is the limiting reactant, it is fully used up in the reaction and there is no Ca remaining after the reaction is completed.
