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Answer:
Percent yield of SiC is 77.0%.
Explanation:
Balanced reaction: 
Molar mass of SiC = 40.11 g/mol
Molar mass of 
= 60.08 g/mol
So, 100.0 kg of = 
moles of = 1664 moles of
According to balanced equation, 1 mol of produces 1 mol of SiC
Therefore, 1664 moles of produce 1664 moles of SiC
Mass of 1664 moles of SiC = 
= 66743g = 66.74 kg (4 sig. fig.)
Percent yield of SiC = [(actual yield of SiC)/(theoretical yield of SiC)]
%
= 
%
= 77.0%
Answer:
3 will be the correct coefficient of CaBr2
Explanation:
In balancing a chemical equation, numbers should be assigned to both reactants and products as a numerical coefficients until all atoms of elements in both sides of the equation count equal.
The balanced equation of the reaction will be:
3CaSO4 + 2AlBr3 ==> 3CaBr2 + Al2(SO4)3
Looking at the unbalanced equation in the question, in the product Al2(SO4)3 there are 3 SO4 group. This will warrant putting 3 behind CaSO4 in order to balance the atoms of SO4 group. That operation will automatically put the number of Ca atoms in CaSO4 to be 3 therefore making CaBr2 to have 3 coefficient as in the balanced equation. This is to balance the number of Ca atoms in both sides to be 3.
The average molecular weight of the mixture can be calculated using this formula:
MWav = x1MW1 + x2MW2
Where x is the mass fraction of the components of the mixture, in this case, copper (63.546 g/mol) and zinc (<span>65.38 g/mol).
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x1 = 7.36 / (7.36+0.51)=0.935
x2 = 0.51 / (7.36+0.51)=0.065
So,
MWav = 0.935(63.546) + 0.065(65.38) = 63.665 g/mol
