<span>substances consisting of ions held together by electrostatic attraction
</span>
We are given that the balanced chemical reaction is:
cacl2⋅2h2o(aq) +
k2c2o4⋅h2o(aq) --->
cac2o4⋅h2o(s) +
2kcl(aq) + 2h2o(l)
We known that
the product was oven dried, therefore the mass of 0.333 g pertains only to that
of the substance cac2o4⋅h2o(s). So what we will do first is to convert this
into moles by dividing the mass with the molar mass. The molar mass of cac2o4⋅h2o(s) is
molar mass of cac2o4 plus the
molar mass of h2o.
molar mass cac2o4⋅h2o(s) = 128.10
+ 18 = 146.10 g /mole
moles cac2o4⋅h2o(s) =
0.333 / 146.10 = 2.28 x 10^-3 moles
Looking at
the balanced chemical reaction, the ratio of cac2o4⋅h2o(s) and k2c2o4⋅h2o(aq) is
1:1, therefore:
moles k2c2o4⋅h2o(aq) = 2.28
x 10^-3 moles
Converting
this to mass:
mass k2c2o4⋅h2o(aq) = 2.28
x 10^-3 moles (184.24 g /mol) = 0.419931006 g
Therefore:
The mass of k2c2o4⋅<span>h2o(aq) in
the salt mixture is about 0.420 g</span>
When looking down the groups the elements have an equal number of electrons in the outer shell. This means they react in a similar way making it easier for scientists to use.
Answer:
2.9 grams.
Explanation:
- From the balanced reaction:
<em>Mg + 1/2O₂ → MgO,</em>
1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.
- We need to calculate the no. of moles of (1.8 g) of Mg and (6.0 g) of oxygen:
no. of moles of Mg = mass/molar mass = (1.8 g)/(24.3 g/mol) = 0.074 mol.
no. of moles of O₂ = mass/molar mass = (6.0 g)/(16.0 g/mol) = 0.375 mol.
<em>So. 0.074 mol of Mg reacts completely with (0.074/2 = 0.037 mol) of O₂ which be in excess.</em>
<em></em>
<em><u>Using cross multiplication:</u></em>
1.0 mole of Mg produce → 1.0 mol of MgO.
∴ 0.074 mol of Mg produce → 0.074 mol of MgO.
<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.074 mol)(40.3 g/mol) = <em>2.98 g.</em>
