Answer: 16.32 g of 
as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of 
require = 1 mole of
Thus 0.34 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of 
Thus 16.32 g of as excess reagent are left.
Answer:
d is the answer of this question
Answer:
= 67.79 g
Explanation:
The equation for the reaction is;
4Cr(s)+3O2(g)→2Cr2O3(s)
The mass of O2 is 21.4 g, therefore, we find the number of moles of O2;
moles O2 = 21.4 g / 32 g/mol
=0.669 moles
Using mole ratio, we get the moles of Cr2O3;
moles Cr2O3 = 0.669 x 2/3
=0.446 moles
but molar mass of Cr2O3 is 151.99 g/mol
Hence,
The mass Cr2O3 = 0.446 mol x 151.99 g/mol
<u> = 67.79 g
</u>
Answer:
1.2,dibromoethane is the sha'awa .
Calcium bicarbonate - Ca(HCO3)2
sodium peroxide - Na202
water - H20
silver nitrate - AgN03
potassium carbonate - K2CO3
sodium carbonate - Na2CO3
zinc chloride - ZnC12
calcium hydroxide - Ca(OH)2
magnesium chloride - MgC12
