2C₂H₂ + 5O₂ = 4CO₂ + 2H₂O
m(CO₂)/{4M(CO₂)} = m(C₂H₂)/{2M(C₂H₂)}
m(CO₂)=2M(CO₂)m(C₂H₂)/M(C₂H₂)
m(CO₂)=2*44g/mol*38.9g/26g/mol = 131.7 g
131.7 grams of carbon dioxide would be formed
Answer:
Sn₃(PO₄)₄ - tin(IV) phosphate.
Explanation:
Hope it helps! :)
27g of 's × 1 mole/32.065g equals 0.84
13.4g of o × 1 mole/16g equals 0.83
Divide all these by the smallest number then round to the nearest number.
When 0.5 moles of Zn react with 0.4 moles of HCL the limiting reactant will be HCL and excess reactant is Zn 0.3 moles excess and 0.2 moles Zn will be consumed in the reaction.
<h3>What is the limiting reactant in the given equation ?</h3>
STEP1- Write the balanced equation
Zn + 2HCL → ZnCl₂ + H₂
STEP2- calculate the moles according to the reaction
1mol of Zn requires → 2moles of HCL
0.5 moles of Zn requires→ 1mol of HCL
Given moles of HCL is 0.4
STEP3- Apply unitary method
0.5/x=1/0.4
0.5×0.4= x
0.2=x
STEP4- excess reagent left 0.5-0.2=0.3
Hence, the 0.3mole of excess Zn is left
Learn more about limiting reactions numericals.
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After the first half-life, 1/2 is unchanged. After the 2nd one, 1/4 remains. After the 4th one, 1/16 remains. So you have 4 half-lives in 48 hours, the half-life is thus 12 hours.
