(1) Dilute HCL is reacted with 4.5 moles of calcium carbonate. Give the equation for the said reaction. Calculate 1) the mass of 4.5 moles of CaCo3. ii) the volume of CO2 liberated at stp. iii) The mass of CaCl2 formed. iv) the number of moles of the acid HCL used in the reaction.
2011 1. Calculate the mass of- i) 10 to the power of 22 atoms of sulphur. ii) 0.1 mole of carbon dioxide. [S=32, C=12 and O=16 and Avogadro number= 6x 10 to the power of 23]
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CaCO3+2HCl = CaCl2 + CO2 + H2O
100 + 73 = 111 + 44 + 18
1) CaCO3 is of mass 450 a.m.u.
4) HCl has 9 moles .
3) CaCl2 is of mass 499.5 a.m.u.
2) CO2 at STP has 100.8 dm power of 3 .
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Answer:
4.5 moles of CaCO3=4.5×100=450
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