Someone plz solve this numerical.
Attachments:
Answers
Answered by
0
baby plz...unblock kro....m wait..
Answered by
1
Heya !
Al₄C₃ + 12H₂O → 4Al(OH)₃ + 2CH₄
144g(Al₄C₃) + 216g( H₂O) → 192g(Al(OH)₃) + 48g(CH₄) { AT STP }
144 g of Al₄C₃ forms 192 g of Al(OH)₃
1 g of Al₄C₃ forms 192/144 g of Al(OH)₃
12 g of Al₄C₃ forms 16 g of Al(OH)₃
144 g of Al₄C₃ forms 22.4 litres of CH₄ at STP
1 g of Al₄C₃ forms 22.4/144 litres of CH₄ at STP
12 g of Al₄C₃ forms 1.8 litres of CH₄ at STP
Al₄C₃ + 12H₂O → 4Al(OH)₃ + 2CH₄
144g(Al₄C₃) + 216g( H₂O) → 192g(Al(OH)₃) + 48g(CH₄) { AT STP }
144 g of Al₄C₃ forms 192 g of Al(OH)₃
1 g of Al₄C₃ forms 192/144 g of Al(OH)₃
12 g of Al₄C₃ forms 16 g of Al(OH)₃
144 g of Al₄C₃ forms 22.4 litres of CH₄ at STP
1 g of Al₄C₃ forms 22.4/144 litres of CH₄ at STP
12 g of Al₄C₃ forms 1.8 litres of CH₄ at STP
Similar questions