So scientist around the world can share there info without having to convert and doing extra work.
Answer:
1.) AgNO₃
2.) 0.563 moles AgBr
Explanation:
The limiting reagent is the reagent that is used up completely during a reaction. It can be identified by calculating which reactant produces the smallest amount of product. This can be done by determining the number of moles of each reagent (via molarity conversion). and then converting it to moles of the product (via mole-to-mole ratio).
AgNO₃ (aq) + KBr (aq) ---> AgBr (s) + KNO₃ (aq)
Molarity (M) = moles / liters
100 mL = 1 L
AgNO₃
45.0 mL / 100 = 45.0 L
1.25 M = ? moles / 0.450 L
? moles = 0.563 moles
KBr
75.0 mL / 100 = 0.750 L
0.800 M = ? moles / 0.750 L
? moles = 0.600 moles
In this case, there is no need to use the mole-to-mole ratio because all of the coefficients are one in the reaction (the amount of the limiting reagent used is the same amount of product produced). Since AgNO₃ produces the smaller amount of product, it is the limiting reagent.
Answer:
2Ba₃(PO₄)₂ +6SiO₂ ⇒ P₄O₁₀ +6BaSiO₃
Explanation:
Equating coefficients, you get ...
aBa₃(PO₄)₂ +bSiO₂ ⇒ cP₄O₁₀ +dBaSiO₃
For Ba: 3a = d
For P: 2a = 4c
For O: 8a +2b = 10c +3d
For Si: b = d
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Expressing everything in terms of b and c, we get ...
d = b
a = b/3 = 2c
From the second, b = 6c, so we have ...
a = 2c
b = 6c
c = c
d = 6c
And we can write the equation with c=1 as ...
2Ba₃(PO₄)₂ +6SiO₂ ⇒ P₄O₁₀ +6BaSiO₃
Heat= mass * change in temperature* specific heat
specific heat=409 J/kg K
