<em>mC₃H₈: 44 g/mol</em>
<em>mCO₂: 44 g/mol</em>
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C₃H₈ + 5O₂ ----> 3CO₂ + 4H₂O
44g (44·3)g
44g C₃H₈ ------ 132g CO₂
15g C₃H₈ ------ X
X = (15×132)/44
<u>X = 45g CO₂
</u>
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First, we apply the law of conservation of mass which states that the total mass in a system remains constant.
Therefore, there must be 5.00 g of sulfur and 4.99 g of oxygen in the product. Now, we determine the mass percentage using:
Mass % = (mass of sulfur x 100) / total mass of compound
Mass % = (5 * 100) / (5 + 4.99)
Mass % = 50.05%
The product contains 50.05% sulfur by mass.
I got B. Because the answer is K2SO4, I got it right on my test so I know its right
Answer is: <span>the mass of the glucose is 81,07 grams.
</span>c(C₆H₁₂O₆) = 0,3 M = 0,3 mol/L.
V(C₆H₁₂O₆) = 1,500 L.
n(C₆H₁₂O₆) = c(C₆H₁₂O₆) · V(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 0,3 mol/L · 1,5 L.
n(C₆H₁₂O₆) = 0,45 mol.
m(C₆H₁₂O₆) = n(C₆H₁₂O₆) · M(C₆H₁₂O₆).
m(C₆H₁₂O₆) = 0,45 mol · 180,156 g/mol.
m(C₆H₁₂O₆) = 81,07 g.
