What is the amount of HCl required to completely react with 200g of CaCO3 according to the equation CaCO3+2HCl--->CaCl2+CO2+H2O
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Answered by
16
For us to work out this question, we need to get the mole ratio from the equation above.
The mole ratio is: 1 :2
That is for every mole of calcium carbonate reacting, there are two moles of HCl reacting.
Molar mass of calcium carbonate :
40(calcium) + 16(oxygen) ×3 + 12(carbon)=100g
100g=1mole
200g=?
200/100 ×1= 2moles.
USING MOLE RATIO TO GET REACTING MOLES OF HCl:
Molar mass of HCl=35.5(Chlorine) + 1(Hydrogen)=36.5g
Mole ratio=1:2
Moles of HCl reacting=2×2=4moles
1 mole=36.5g
4moles =?
4×36.5=146grams
The mass of HCl reacting with 200g of calcium carbonate is 146 grams.
The mole ratio is: 1 :2
That is for every mole of calcium carbonate reacting, there are two moles of HCl reacting.
Molar mass of calcium carbonate :
40(calcium) + 16(oxygen) ×3 + 12(carbon)=100g
100g=1mole
200g=?
200/100 ×1= 2moles.
USING MOLE RATIO TO GET REACTING MOLES OF HCl:
Molar mass of HCl=35.5(Chlorine) + 1(Hydrogen)=36.5g
Mole ratio=1:2
Moles of HCl reacting=2×2=4moles
1 mole=36.5g
4moles =?
4×36.5=146grams
The mass of HCl reacting with 200g of calcium carbonate is 146 grams.
Answered by
4
Answer:
146 grams
Explanation:
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