1)we need a balanced equation: CuSO₄ + Zn ---> ZnSO₄ + Cu
2) we need to convert the grams of CuSO₄ to moles using the molar mass.
molar mass CuSO₄= 63.5 + 32.0 + (4 x 16.0)= 160 g/mol
3) convert moles of CuSO₄ to moles of Cu
4) convert moles of Cu to grams using it's molar mass.
molar mass Cu= 63.5 g/mol
I did it step-by-step as the explanation but you can do all of this in one step.
Full Question:
A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?
How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?
2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)
Answer:
13.1 g K2CO3 required to neutralize spill
Explanation:
2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)
Number of moles = Volume * Molar Concentration
moles HBr= 0.42L x .45 M= 0.189 moles HBr
From the stoichiometry of the reaction;
1 mole of K2CO3 reacts with 2 moles of HBr
1 mole = 2 mole
x mole = 0.189
x = 0.189 / 2 = 0.0945 moles
Mass = Number of moles * Molar mass
Mass = 0.0945 * 138.21 = 13.1 g
Hello!
Calvin told Marie that they could continue to add solute until the reached 40 grams because the solution was still unsaturated.
Unsaturated solutions are those in which the solvent (in this case water) can still dissolve more solute (in this case KNO₃) at the given pressure and temperature. This can be seen visually when adding more solute doesn't result in the presence of grains of solids that settle in the bottom of the flask. That happens because the rate of dissolving is higher than the rate of crystallization.
Have a nice day!
The balanced equation for combustion is as follows;
2CH₃OH + 3O₂ ---> 2CO₂ + 4H₂O
The stoichiometry of CH₃OH to O₂ is 2:3
the limiting reagent is the reactant that is fully consumed during the reaction. The amount of product formed is directly proportional to the amount of limiting reactant produced. The excess reagent is the reactant that is provided in excess and is not fully used up, there will be an amount of this reagent remaining after the reaction.
If methanol is the limiting reactant,
If 2 mol of methanol reacts with 3 moles of O₂
Then 24 mol of methanol reacts with - 3/2 x 24 = 36 mol of O₂ should be present
But only 15 mol of O₂ is present, therefore O₂ is the limiting reactant and methanol is in excess.
3 mol of O₂ reacts with 2 mol of CH₃OH
then 15 mol of O₂ reacts with 2/3 x 15 = 10 mol of CH₃OH
Excess reactant is methanol, 10 mol are used up therefore 24 - 10 mol = 14 mol are remaining at the end of the reaction
