Mass of strontium = 88 g
Atomic mass of Sr = 87.62 u
Moles = mass /atomic mass
Moles of Sr = 88 / 87.62 = 1.004
Mass of bromine = 160 g
Atomic mass of Br = 79.904 u
Molar mass of Br₂ = 79.904 x 2 = 159.808 g/mol
Moles = mass /molar mass
Moles of Br₂ = 160 / 159.808 = 1.001
The balanced chemical equation of the reaction between strontium and bromine:
Sr(s) + Br₂(l) = SrBr₂
The molar ratio between Br₂ and SrBr₂ is 1:1
So the moles of SrBr₂ produced from 1.001 moles of Br₂ is 1.001.
Moles = mass /molar mass
Mass = moles x molar mass
Molar mass of SrBr₂ = 247.4280 g/mol
Mass = 1.001 mol x 247.4280 g/mol
Mass = 247.428 g
The mass of strontium bromide produced is 247.428 g.
First step is to calculate the mole of each element
that is;
carbon 97.6/12=8.13moles
hydrogen= 4.9/1=4.9 moles
oxygen 52/16=3.25 moles
nitrogen=45.5/14=3.25 moles
step two is to calculate the mole ratio by dividing with the smallest number of moles
that is divide each mole with 3.25moles
carbon=8.13/3.25 =5/2
hydrogen=4.9/3.25= 3/2
oxygen=3.25/3.25=1
nitrogen=3.25/3.25=1
step 3; multiply all the mole ratio by 2 to remove the fraction
carbon=5/2 x2 =5
hydrogen=3/2 x2=3
oxygen=1 x2=2
nitrogen =1x2=2
therefore the empirical formula is C5H3O2N2(answer c)
4 l ------ 3,2 g
x l ------ 8 g
x = 8 g × 4 l / 3,2 g = 10 l
Answer: 10 l of sulfuric acid is needed to produce 8 g of product.
:-) ;-)
D. this is the correct answer because the question is showing a comparison of sizes rather than the actual sizes which eliminates B. A and C have no relevance.
