Dear sir,
i am currently a university student Studying Engineering in Bangladesh. i have completed my O level and A level under Edexcel IGCSE and GCE Edexcel which are under London curriculum and so i am well aware of excellent English. I am dedicated to provide you services at your requirement and i know that i have the ability to make you satisfied.
Please do Find the Answer for the first question that you have mentioned in the pdf file.
Ans1: a) Na2 CO3+ 2HCl →2NaCl+CO_2+ H_2 O
b) Mole= Volume/(24000 〖ml〗^3 )
=1871/24000
=0.07795 mol
=0.078 mol
Mole Ratio of CO2: HCl = 1 : 2
So, the number of mole of HCl is 2 x 0.078 mol.
= 0.156 mol
The mole ratio of HCl : Na2 CO3 = 2 : 1
So, the number of mole of Na2 CO3 is 0.156/2
= 0.078 mol
Mass= mole x Molar mass
= 0.078 x [(23x2) + 12 + (16x3)]
= 8.268 g
= 8.27 g
Percentage of Na2 CO3 is 8.27/10 x 100 %
= 82.7 %
Hope to hear a positive and prompt reply from your end.
Thank you
Regards
Ashir