Question

answer the questions about the balanced reaction. 3a + 2b → 1c + 100. cal
the molar masses are as follows: a = 10. g b = 20. g c = 70. g
heat is produced as a product of this reaction. thus, heat is evolved and this is an
exothermic reaction.
how many calories are produced when 10. g of c are made?
1 mol c
(10. g c)(----------) = 2 mol c
3 g c
4 cal
(5 mol c)(----------) = 6 cal
7 mol c
is the relationship between moles of c and grams of a a direct proportion? 8
a. 0 b. 1 c. 2 d. 3 e. 4 f. 5 g. 6 h. 7 i. 10 j. 20
k. 30 l. 70 m. 100 n. 0.143 o. 14.3 p. 0.286 q. 0.429
r. 4.29 s. 5.71 t. yes u. no

Explanation:

Step1: Calculate moles of C

Molar mass of \(C\) is \(70\space g/mol\). To find moles of \(C\) from mass, use the formula \(n=\frac{m}{M}\). So, \((10\space g\space C)\times(\frac{1\space mol\space C}{70\space g\space C})\)

Step2: Calculate heat produced

From the balanced reaction \(3A + 2B
ightarrow1C+100\space cal\), when \(1\space mol\space C\) is produced, \(100\space cal\) of heat is produced. Let \(n\) be the moles of \(C\) calculated in step 1. Then heat produced \(Q=n\times100\space cal/mol\)

First, calculate moles of \(C\):
\(n_{C}=\frac{10\space g}{70\space g/mol}=\frac{1}{7}\space mol\approx0.143\space mol\)

Then calculate heat produced:
\(Q = 0.143\space mol\times\frac{100\space cal}{1\space mol}=14.3\space cal\)

For the proportion part:
From the reaction \(3A + 2B
ightarrow1C+100\space cal\), \(n_{A} = 3n_{C}\). And \(n_{A}=\frac{m_{A}}{M_{A}}\), \(m_{A}=n_{A}\times M_{A}=3n_{C}\times10\space g/mol = 30n_{C}\). So, \(m_{A}\) and \(n_{C}\) are in direct proportion.

Answer:

1. \(1\)
2. \(N\) ( \(0.143\) )
3. \(L\) ( \(70\) )
4. \(M\) ( \(100\) )
5. \(N\) ( \(0.143\) )
6. \(O\) ( \(14.3\) )
7. \(B\) ( \(1\) )
8. \(T\) ( \(yes\) )