zadanie 3 na podstawie równania reakcji: 2Ca+O2->2CaO oblicz:
a)ile moli wapnia potrzeba aby otrzymać 8 moli tlenku wapnia
b)ile dm(trójka na górze)tlenu należy użyć aby utlenić 3 mole wapnia.
c)ile gram tlenku wapnia powstanie z 1 mola wapnia.
Ps-proszę o szybką pomoc i rozwiązanie daje 25 punktów za rozwiązanie pozdrawiam.
Izu123456
1. a) mCa= 40g 1 mol Ca --- 40g 3 mol Ca --- x x = 3 *40g x = 120g
b) mCO₂= 12g + 2 * 16g = 44g 1 mol CO₂ --- 44g 5 mol CO₂ --- x x = 5*44g x = 220g
2. a) mCH₄= 12g + 4* 1g = 16g 1 mol CH₄ --- 16g x --- 100g 16x = 100 /: 16 x = 6,25 mol
b) mNaOH= 23g + 16g + 1g = 40g 1 mol NaOH --- 40g x --- 50g 40x = 50 /:40 x = 1,25 mol
3. a) 2 mol Ca --- 2 mol CaO x --- 8 mol CaO 2x = 2 * 8 /:2 x= 8 mol Ca
b) 2 mol Ca --- 2 * 22,4 dm³ 3 mol Ca --- x 2x = 3 * 2 * 22,4 /:2 x= 67,2 dm³
c) m CaO = 40g + 16g = 56 g 2 mol Ca --- 56 g CaO 1 mol Ca --- x 2x = 56g /:2 x = 28 g
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GniewnaDiablica
1. a) 1 mol Ca - 40 g Ca 3 mole Ca - x g Ca x= 40*3=120 g Ca b) 1 mol CO2 - (2*16+12) 44 g CO2 5 moli CO2 - x g CO2 x = 5*44 = 220 g CO2
2. a) 100 g CH4 - x moli CH4 16 g (12+4*1) CH4 - 1 mol CH4 x= 100:16 = 6,25 mola CH4
b) 50 g NaOH - x moli NaOH 40 g (23+16+1) - 1 mol NaOH x = 50:40 = 1,25 mola NaOH
2. a) 2 mole Ca - 2 mole CaO x moli Ca - 8 moli CaO x = (8*2):2 = 8 moli CaO b) 22,4 dm3 O - 1 mol Ca x dm3 O - 3 mole Ca x = 3*22,4 = 67,2 dm3 O c) 2 mole Ca - 96g CuO 1 mol Ca - x g CuO x = 96:2 =48 g CuO
a) mCa= 40g
1 mol Ca --- 40g
3 mol Ca --- x
x = 3 *40g
x = 120g
b) mCO₂= 12g + 2 * 16g = 44g
1 mol CO₂ --- 44g
5 mol CO₂ --- x
x = 5*44g
x = 220g
2.
a) mCH₄= 12g + 4* 1g = 16g
1 mol CH₄ --- 16g
x --- 100g
16x = 100 /: 16
x = 6,25 mol
b) mNaOH= 23g + 16g + 1g = 40g
1 mol NaOH --- 40g
x --- 50g
40x = 50 /:40
x = 1,25 mol
3.
a) 2 mol Ca --- 2 mol CaO
x --- 8 mol CaO
2x = 2 * 8 /:2
x= 8 mol Ca
b) 2 mol Ca --- 2 * 22,4 dm³
3 mol Ca --- x
2x = 3 * 2 * 22,4 /:2
x= 67,2 dm³
c) m CaO = 40g + 16g = 56 g
2 mol Ca --- 56 g CaO
1 mol Ca --- x
2x = 56g /:2
x = 28 g
a)
1 mol Ca - 40 g Ca
3 mole Ca - x g Ca
x= 40*3=120 g Ca
b)
1 mol CO2 - (2*16+12) 44 g CO2
5 moli CO2 - x g CO2
x = 5*44 = 220 g CO2
2.
a)
100 g CH4 - x moli CH4
16 g (12+4*1) CH4 - 1 mol CH4
x= 100:16 = 6,25 mola CH4
b)
50 g NaOH - x moli NaOH
40 g (23+16+1) - 1 mol NaOH
x = 50:40 = 1,25 mola NaOH
2.
a)
2 mole Ca - 2 mole CaO
x moli Ca - 8 moli CaO
x = (8*2):2 = 8 moli CaO
b)
22,4 dm3 O - 1 mol Ca
x dm3 O - 3 mole Ca
x = 3*22,4 = 67,2 dm3 O
c)
2 mole Ca - 96g CuO
1 mol Ca - x g CuO
x = 96:2 =48 g CuO